1. Show that the sum of $\displaystyle (m+n)^{th}$ and $\displaystyle (m-n)^{th}$ terms of an A.P. is equal to twice the $\displaystyle m^{th}$ term.
Let a be the first term and d be the common difference of A.P.
We know that $\displaystyle k^{th}$ term of an A.P. is $\displaystyle a_{k}=a+(k-1)d$
⇒ $\displaystyle a_{m+n}=a+(m+n-1)d$ and $\displaystyle a_{m-n}=a+(m-n-1)d$
$\displaystyle a_{m+n}+a_{m-n}$ $\displaystyle =a+(m+n-1)d+a$ $\displaystyle +(m-n-1)d$
$\displaystyle =2a+(m+n-1+m-n-1)d$
$\displaystyle =2a+(2m-2)d$
$\displaystyle =2[a+(m-1)d]$
$\displaystyle =2a_{m}$
Thus sum of $\displaystyle (m+n)^{th}$ and $\displaystyle (m-n)^{th}$ terms is equal to twice the $\displaystyle m^{th}$ term.
2. If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.
Let the three numbers be a-d, a and a+d
Given that,
$\displaystyle (a-d)+a+(a+d)=24$
$\displaystyle 3a=24$
$\displaystyle a=8$
Also given that,
$\displaystyle (a-d)a(a+d)=440$
since a=8,
$\displaystyle (8-d)8(8+d)=440$
$\displaystyle (8-d)(8+d)=55$
$\displaystyle 8^{2}-d^{2}=55$
$\displaystyle 64-d^{2}=55$
$\displaystyle d^{2}=64-55=9$
$\displaystyle d=\pm 3$
When d=3, the numbers are 5, 8, 11 and when d=-3 the numbers are 11, 8, 5.
Hence, the numbers are 5, 8 and 11
3. Let the sum of n, 2n, 3n terms of an A.P. be $\displaystyle S_{1},S_{2},S_{3}$ respectively, show that $\displaystyle S_{3}=3(S_{2}-S_{1})$
Let a be the first term and d be the common difference of A.P.
$\displaystyle S_{1}=\frac{n}{2}(2a+(n-1)d)$
$\displaystyle S_{2}=\frac{2n}{2}(2a+(2n-1)d)$
$\displaystyle S_{1}=\frac{3n}{2}(2a+(3n-1)d)$
Consider $\displaystyle 3(S_{2}-S_{1})$
$\displaystyle =3\left [ \frac{2n}{2}(2a+(2n-1)d) -\frac{n}{2}(2a+(n-1)d)\right ]$
$\displaystyle =\frac{3n}{2}\left [ 2(2a+(2n-1)d) -(2a+(n-1)d)\right ]$
$\displaystyle =\frac{3n}{2}\left [ (4a+4nd-2d) -(2a+nd-d)\right ]$
$\displaystyle =\frac{3n}{2}\left [ (4a+4nd-2d-2a-nd+d)\right ]$
$\displaystyle =\frac{3n}{2}(2a+3nd-d)$
$\displaystyle =\frac{3n}{2}(2a+(3n-1)d)=S_{3}$
4. Find the sum of all numbers between 200 and 400 which are divisible by 7.
The numbers between 200 and 400 which are divisible by 7 are 203, 210,...,399.
Therefore, a=203
l=399
d=7
Let the numbers of terms be n
$\displaystyle a_{n}=a+(n-1)d=399$
$\displaystyle 203+(n-1)7=399$
$\displaystyle (n-1)7=399-203=196$
$\displaystyle (n-1)=\frac{196}{7}=28$
$\displaystyle n=29$
$\displaystyle S_{n}=\frac{n}{2}(a+l)$
$\displaystyle =\frac{29}{2}(203+399)$
$\displaystyle =\frac{29}{2}(602)=8729$
5. Find the sum of integers from 1 to 100 that are divisible by 2 or 5.
The integers from 1 to 100 that are divisible by 2 are 2,4,..100.
Therefore, a=2, l=100, d=2
$\displaystyle a_{n}=a+(n-1)d$
$\displaystyle 100=2+(n-1)d$
$\displaystyle 100-2=(n-1)2$
$\displaystyle 98=(n-1)2$
$\displaystyle n-1=49$ ⇒ $\displaystyle n=50$
$\displaystyle S_{n}=\frac{n}{2}(a+l)$
$\displaystyle =\frac{50}{2}(2+100)$
$\displaystyle =25(102)=2550$
The integers from 1 to 100 that are divisible by 5 are 5,10,..100.
Therefore, a=5, l=100, d=5
$\displaystyle a_{n}=a+(n-1)d$
$\displaystyle 100=5+(n-1)5$
$\displaystyle 100-5=(n-1)5$
$\displaystyle 95=(n-1)5$
$\displaystyle n-1=19$ ⇒ $\displaystyle n=20$
$\displaystyle S_{n}=\frac{n}{2}(a+l)$
$\displaystyle =\frac{20}{2}(5+100)$
$\displaystyle =10(105)=1050$
The integers from 1 to 100 that are divisible by 2 and 5 are 10,20,..100.
Therefore, a=10, l=100, d=10
$\displaystyle a_{n}=a+(n-1)d$
$\displaystyle 100=10+(n-1)10$
$\displaystyle 100-10=(n-1)10$
$\displaystyle 90=(n-1)10$
$\displaystyle n-1=9$ ⇒ $\displaystyle n=10$
$\displaystyle S_{n}=\frac{n}{2}(a+l)$
$\displaystyle =\frac{10}{2}(10+100)$
$\displaystyle =5(110)=550$
Therefore the required sum is $\displaystyle 2550+1050-550=3050$
6. Find the sum of all two digit numbers which when divided by 4 yields 1 as remainder.
Two digit numbers which when divided by 4 yields 1 as remainder are 13,17,...97
Therefore, a=13, l=97, d=4
$\displaystyle a_{n}=a+(n-1)d$
$\displaystyle 97=13+(n-1)4$
$\displaystyle 97-13=(n-1)4$
$\displaystyle 84=(n-1)4$
$\displaystyle n-1=21$ ⇒ $\displaystyle n=22$
$\displaystyle S_{n}=\frac{n}{2}(a+l)$
$\displaystyle =\frac{22}{2}(13+97)$
$\displaystyle =11(110)=1210$
7. If f is a function satisfying $\displaystyle f(x+y)=f(x)f(y)$ for all $\displaystyle x,y\epsilon N$ such that f(1)=3 and $\displaystyle \sum_{x=1}^{n}f(x)=120$, find the value of n.
Given that, $\displaystyle f(x+y)=f(x)f(y)$ and f(1)=3
Taking x=y=1,
$\displaystyle f(1+1)=f(1)f(1)$
$\displaystyle f(2)=3\times 3=3^{2}=9$
Similarly,
$\displaystyle f(3)=3^{3}=27$
$\displaystyle f(4)=3^{4}=81$
f(1), f(2), f(3),... i.e. 3,9,27,... forms a G.P.
⇒ a=3, r=3
$\displaystyle \sum_{x=1}^{n}f(x)=120$
$\displaystyle f(1)+f(2)+f(3)+...+f(n)=120$
$\displaystyle 3+3^{2}+3^{3}+...+3^{n}=120$
$\displaystyle \frac{3(3^{n}-1)}{3-1}=120$
$\displaystyle \frac{3(3^{n}-1)}{2}=120$
$\displaystyle 3(3^{n}-1)=240$
$\displaystyle 3^{n}-1=80$
$\displaystyle 3^{n}=81$
⇒ $\displaystyle 3^{n}=3^{4}$
⇒ n=4
8. The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2 respectively. Find the last term and the number of terms.
Given that, a=5, r=2
Let sum of n terms be 315.
$\displaystyle S_{n}=\frac{a(r^{n}-1)}{r-1}=315$
$\displaystyle \frac{5(2^{n}-1)}{2-1}=315$
$\displaystyle 5(2^{n}-1)=315$
$\displaystyle 2^{n}-1=63$
$\displaystyle 2^{n}=64=2^{6}$
Therefore, number of terms n=6
Last term = $\displaystyle ar^{n-1}$
$\displaystyle 5(2)^{6-1}$
$\displaystyle 5(2)^{5}=5\times 32=160$
9. The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.
Given that, a=1,
$\displaystyle a_{3}+a_{5}=90$
$\displaystyle ar^{2}+ar^{4}=90$
$\displaystyle r^{2}+r^{4}=90$
$\displaystyle r^{4}+r^{2}-90=0$
$\displaystyle r^{2}=\frac{1\pm \sqrt{1+4\times 90}}{2}$
$\displaystyle r^{2}=\frac{1\pm \sqrt{1+360}}{2}$
$\displaystyle r^{2}=\frac{1\pm \sqrt{361}}{2}$ $\displaystyle =\frac{1\pm 19}{2}$ $\displaystyle =-10 or 9$
$\displaystyle r=\pm 3$
10. The sum of three numbers in G.P. is 56. If we subtract 1,7,21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.
Let the numbers be $\displaystyle a,ar,ar^{2}$
Given that, $\displaystyle a+ar+ar^{2}=56$ ...(i)
and $\displaystyle a-1, ar-7,ar^{2}-21$ form an A.P.
⇒ $\displaystyle (ar-7)-(a-1)=(ar^{2}-21)-(ar-7)$
$\displaystyle ar-a-6=ar^{2}-ar-14$
$\displaystyle ar^{2}-2ar+a=8$
$\displaystyle a(r^{2}-2r+1)=8$ ...(ii)
$\displaystyle \frac{(i)}{(ii)}=\frac{a(1+r+r^{2})}{a(r^{2}-2r+1)}=\frac{56}{8}$
$\displaystyle \frac{1+r+r^{2}}{r^{2}-2r+1}=7$
$\displaystyle 1+r+r^{2}=7(r^{2}-2r+1)$
$\displaystyle 6r^{2}-15r+6=0$
$\displaystyle 2r^{2}-5r+2=0$
$\displaystyle (r-2)(2r-1)=0$
⇒ $\displaystyle r=2\: or\: \frac{1}{2}$
when $\displaystyle r=2$ , (i) ⇒ a=8 and three numbers are 8,16,32
when $\displaystyle r=\frac{1}{2}$ , (i) ⇒ a=32 and three numbers are 32,16,8
11. A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.
Let the number of terms be 2n ⇒ G.P. is $\displaystyle G_{1},G_{2},G_{3}...G_{2n}$
Given that, $\displaystyle G_{1}+G_{2}+G_{3}...+G_{2n}=5[G_{1}+G_{3}...+G_{2n-1}]$
$\displaystyle G_{1}+G_{2}+G_{3}...+G_{2n}-5[G_{1}+G_{3}...+G_{2n-1}]=0$
$\displaystyle G_{2}+G_{4}...+G_{2n}-4[G_{1}+G_{3}...+G_{2n-1}]=0$
$\displaystyle G_{2}+G_{4}...+G_{2n}=4[G_{1}+G_{3}...+G_{2n-1}]$
$\displaystyle ar+ar^{3}+...ar^{2n-1}=4[a+ar^{2}+...ar^{2n-2}]$
⇒ r=4
12. The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.
Let the first four terms are a,a+d,a+2d,a+3d
and the last four terms are a+(n-4)d,a+(n-3)d,a+(n-2)d,a+(n-1)d
Given that,
$\displaystyle a+a+d+a+2d+a+3d=56$
$\displaystyle 4a+6d=56$
$\displaystyle 4(11)+6d=56$
$\displaystyle 44+6d=56$
$\displaystyle 6d=12$ ⇒ $\displaystyle d=2$
Also given that,
$\displaystyle a+(n-4)d+a+(n-3)d+a+(n-2)d+a+(n-1)d=112$
$\displaystyle 4a+(4n-10)d=112$
$\displaystyle 4(11)+(4n-10)2=112$
$\displaystyle 44+8n-20=112$
$\displaystyle 8n=88$ ⇒ $\displaystyle n=11$
13. If $\displaystyle \frac{a+bx}{a-bx}=\frac{b+cx}{b-cx}=\frac{c+dx}{c-dx}$ (xne0), then show that a,b,c and d are in G.P.
Given that,
$\displaystyle \frac{a+bx}{a-bx}=\frac{b+cx}{b-cx}$
$\displaystyle (a+bx)(b+cx)=(b-cx)(a-bx)$
$\displaystyle ab-acx+b^{2}x-bcx^{2}=ab-b^{2}x+acx-bcx^{2}$
$\displaystyle 2b^{2}x=2acx$
⇒ $\displaystyle \frac{b}{c}=\frac{a}{b}$ ...(i)
Also given that,
$\displaystyle \frac{b+cx}{b-cx}=\frac{c+dx}{c-dx}$
$\displaystyle (b+cx)(c+dx)=(c-dx)(b-cx)$
$\displaystyle bc-bdx+c^{2}x-cdx^{2}=bc-c^{2}x+bdx-cdx^{2}$
$\displaystyle 2c^{2}x=2bdx$
⇒ $\displaystyle \frac{c}{d}=\frac{b}{c}$ ...(ii)
from (i) and (ii), $\displaystyle \frac{a}{b}=\frac{b}{c}=\frac{c}{d}$
⇒ a,b,c and are in G.P.
14. Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that $\displaystyle P^{2}R^{n}=S^{n}$
Let the G.P. be $\displaystyle a,ar,ar^{2},...ar^{n-1}$
Given that,
$\displaystyle S=\frac{a(r^{n}-1)}{r-1}$
$\displaystyle P=a^{n}r^{1+2+...n-1}=a^{n}r^{\frac{n(n-1)}{2}}$
$\displaystyle R=\frac{1}{a}+\frac{1}{ar}+...\frac{1}{ar^{n-1}}$
$\displaystyle R=\frac{r^{n-1}+r^{n-2}+...r+1}{ar^{n-1}}$
$\displaystyle =(1+r+...r^{n-2}+r^{n-1})\frac{1}{ar^{n-1}}$
$\displaystyle =\frac{1(r^{n}-1)}{r-1}\times \frac{1}{ar^{n-1}}$
$\displaystyle =\frac{r^{n}-1}{ar^{n-1}(r-1)}$
Therefore,
$\displaystyle P^{2}R^{n}=a^{2n}r^{\frac{2n(n-1)}{2}}\times \frac{(r^{n}-1)^{n}}{a^{n}r^{n(n-1)}(r-1)^{n}}$
$\displaystyle =\frac{a^{2n}r^{n(n-1)}(r^{n}-1)^{n}}{a^{n}r^{n(n-1)}(r-1)^{n}}$
$\displaystyle =\frac{a^{n}(r^{n}-1)^{n}}{(r-1)^{n}}$
$\displaystyle =\left [\frac{a(r^{n}-1)}{r-1} \right ]^{n}=S^{n}$
15. The $\displaystyle p^{th},q^{th},r^{th}$ terms of an A.P. are a,b,c respectively. Show that $\displaystyle (q-r)a+(r-p)b+(p-q)c=0$
Let f be the first term and d be the common difference.
Given that,
$\displaystyle a_{p}=f+(p-1)d=a$ ...(i)
$\displaystyle a_{q}=f+(q-1)d=b$ ...(ii)
$\displaystyle a_{r}=f+(r-1)d=c$ ...(iii)
(i)-(ii) ⇒
$\displaystyle a-b=f+(p-1)d-(f+(q-1)d)$
$\displaystyle a-b=(p-1)d-(q-1)d$
$\displaystyle a-b=(p-1-q+1)d$
$\displaystyle a-b=(p-q)d$ ...(iv)
(ii)-(iii) ⇒
$\displaystyle b-c=f+(q-1)d-(f+(r-1)d)$
$\displaystyle b-c=(q-1)d-(r-1)d$
$\displaystyle b-c=(q-1-r+1)d$
$\displaystyle b-c=(q-r)d$ ...(v)
Equatin (iv)and (v) ⇒
$\displaystyle \frac{a-b}{p-q}=\frac{b-c}{q-r}$
$\displaystyle (a-b)(q-r)=(b-c)(p-q)$
$\displaystyle aq-ar-bq+br=bp-bq-cp+cq$
$\displaystyle aq-ar-bq+br-bp+bq+cp-cq=0$
$\displaystyle a(q-r)+b(-q+r-p+q)+c(p-q)=0$
$\displaystyle a(q-r)+b(r-p)+c(p-q)=0$
16. If $\displaystyle a\left ( \frac{1}{b}+\frac{1}{c} \right )$, $\displaystyle b\left ( \frac{1}{c}+\frac{1}{a} \right )$, $\displaystyle c\left ( \frac{1}{a}+\frac{1}{b} \right )$ are in A.P. prove that a,b,c are in A.P.
Given that,
$\displaystyle a\left ( \frac{1}{b}+\frac{1}{c} \right )$, $\displaystyle b\left ( \frac{1}{c}+\frac{1}{a} \right )$, $\displaystyle c\left ( \frac{1}{a}+\frac{1}{b} \right )$ are in A.P. ⇒
$\displaystyle b\left ( \frac{1}{c}+\frac{1}{a} \right )$-$\displaystyle a\left ( \frac{1}{b}+\frac{1}{c} \right )$=$\displaystyle c\left ( \frac{1}{a}+\frac{1}{b} \right )$-$\displaystyle b\left ( \frac{1}{c}+\frac{1}{a} \right )$
$\displaystyle b\left ( \frac{a+c}{ac} \right )-a\left ( \frac{b+c}{bc} \right )=c\left ( \frac{a+b}{ab} \right )-b\left ( \frac{a+c}{ac} \right )$
$\displaystyle \left ( \frac{ab+bc}{ac} \right )-\left ( \frac{ab+ac}{bc} \right )=\left ( \frac{ac+bc}{ab} \right )-\left ( \frac{ab+bc}{ac} \right )$
$\displaystyle \frac{ab^{2}+b^{2}c-a^{2}b-a^{2}c}{abc} = \frac{ac^{2}+bc^{2}-ab^{2}-b^{2}c}{abc}$
$\displaystyle ab^{2}+b^{2}c-a^{2}b-a^{2}c=ac^{2}+bc^{2}-ab^{2}-b^{2}c$
$\displaystyle ab(b-a)+c(b^{2}-a^{2})=a(c^{2}-b^{2})+bc(c-b)$
$\displaystyle ab(b-a)+c(b-a)(b+a)=a(c-b)(c+b)+bc(c-b)$
$\displaystyle (b-a)(ab+cb+ca)=(c-b)(ac+ab+bc)$
$\displaystyle b-a=c-b$
Thus a,b,c are in A.P.
17. If a,b,c,d are in G.P. prove that $\displaystyle (a^{n}+b^{n}),(b^{n}+c^{n}),(c^{n}+d^{n})$ are in G.P.
Given that a,b,c,d are in G.P.
$\displaystyle \frac{c}{b}=\frac{b}{a}$ ⇒ $\displaystyle b^{2}=ac$ ...(i)
$\displaystyle \frac{d}{c}=\frac{c}{b}$ ⇒ $\displaystyle c^{2}=bd$ ...(ii)
$\displaystyle \frac{b}{a}=\frac{d}{c}$ ⇒ $\displaystyle bc=ad$ ...(iii)
To prove $\displaystyle (a^{n}+b^{n}),(b^{n}+c^{n}),(c^{n}+d^{n})$ are in G.P., let us prove $\displaystyle (b^{n}+c^{n})^{2}=(a^{n}+b^{n})(c^{n}+d^{n})$
Consider L.H.S.
$\displaystyle (b^{n}+c^{n})^{2}=b^{2n}+c^{2n}+2b^{n}c^{n}$
$\displaystyle =(b^{2})^{n}+(c^{2})^{n}+2(bc)^{n}$
$\displaystyle =(b^{2})^{n}+(c^{2})^{n}+(bc)^{n}+(bc)^{n}$
$\displaystyle =(ac)^{n}+(bd)^{n}+(bc)^{n}+(ad)^{n}$ [using (i),(ii),(iii)]
$\displaystyle =a^{n}c^{n}+b^{n}c^{n}+b^{n}d^{n}+a^{n}d^{n}$
$\displaystyle =c^{n}(a^{n}+b^{n})+d^{n}(a^{n}+b^{n})$
$\displaystyle =(a^{n}+b^{n})(c^{n}+d^{n})$
L.H.S.=R.H.S.
18. If a and b are the roots of $\displaystyle x^{2}-3x+p=0$ and c,d are roots of $\displaystyle x^{2}-12x+q=0$ where a,b,c,d form a G.P. Prove that $\displaystyle (q+p):(q-p)=17:15$
Given that, a,b are roots of $\displaystyle x^{2}-3x+p=0$
therefore a+b=3 and ab=p ...(i)
And c,d are roots of $\displaystyle x^{2}-12x+q=0$
therefore c+d=12 and cd=q ...(ii)
Also given that a,b,c,d are in G.P.
Let $\displaystyle a=x,b=xr,c=xr^{2},d=xr^{3}$
From (i) and (ii)
$\displaystyle x+xr=3$ ⇒ $\displaystyle x(1+r)=3$ ...(iii)
$\displaystyle xr^{2}+xr^{3}=12$ ⇒ $\displaystyle xr^{2}(1+r)=12$ ...(iv)
Dividing (iv) by (iii),
$\displaystyle \frac{xr^{2}(1+r)}{x(1+r)}=\frac{12}{3}$
$\displaystyle r^{2}=4$ ⇒ $\displaystyle r=\pm 2$
Case I: when $\displaystyle r=2, x=\frac{3}{1+2}=1$
from (i) and (ii)
$\displaystyle p=ab=x^{2}r=2$
$\displaystyle q=cd=x^{2}r^{5}=32$
$\displaystyle \frac{q+p}{q-p}=\frac{32+2}{32-2}=\frac{34}{30}=\frac{17}{15}$
Case II: when $\displaystyle r=-2, x=\frac{3}{1-2}=-3$
from (i) and (ii)
$\displaystyle p=ab=x^{2}r=-18$
$\displaystyle q=cd=x^{2}r^{5}=-288$
$\displaystyle \frac{q+p}{q-p}=\frac{-288-18}{-288+18}=\frac{-306}{-270}=\frac{17}{15}$
19. The ratio of the A.M. and G.M. of two positive numbers a and b is m:n. Show that $\displaystyle a:b=(m+\sqrt{m^{2}-n^{2}}):(m-\sqrt{m^{2}-n^{2}})$
Consider the numbers a,b
A.M.=$\displaystyle \frac{a+b}{2}$ and G.M.=$\displaystyle \sqrt{ab}$
Given that, $\displaystyle \frac{AM}{GM}=\frac{m}{n}$
$\displaystyle \frac{a+b}{2\sqrt{ab}}=\frac{m}{n}$
$\displaystyle a+b=\frac{2\sqrt{ab}m}{n}$ ...(i)
Consider,
$\displaystyle (a-b)^{2}=(a+b)^{2}-4ab$
$\displaystyle (a-b)^{2}=\left ( \frac{2\sqrt{ab}m}{n} \right )^{2}-4ab$
$\displaystyle (a-b)^{2}=\frac{4abm^{2}}{n^{2}} -4ab$
$\displaystyle (a-b)^{2}=4ab\left ( \frac{m^{2}}{n^{2}} -1 \right )$
$\displaystyle (a-b)^{2}=4ab\left ( \frac{m^{2}-n^{2}}{n^{2}} \right )$
$\displaystyle a-b=2\sqrt{ab}\left ( \frac{\sqrt{m^{2}-n^{2}}}{n} \right )$ ...(ii)
Adding (i) and (ii)
$\displaystyle a+b+a-b=\frac{2\sqrt{ab}m}{n}+2\sqrt{ab}\left ( \frac{\sqrt{m^{2}-n^{2}}}{n} \right )$
$\displaystyle 2a=\frac{2\sqrt{ab}}{n}(m+\sqrt{m^{2}-n^{2}})$
$\displaystyle a=\frac{\sqrt{ab}}{n}(m+\sqrt{m^{2}-n^{2}})$
Substitute value of a in (i) ⇒
$\displaystyle b=\frac{2\sqrt{ab}m}{n}-a$
$\displaystyle b=\frac{2\sqrt{ab}m}{n}-\frac{\sqrt{ab}}{n}(m+\sqrt{m^{2}-n^{2}})$
$\displaystyle b=\frac{\sqrt{ab}}{n}(2m-m-\sqrt{m^{2}-n^{2}})$
$\displaystyle b=\frac{\sqrt{ab}}{n}(m-\sqrt{m^{2}-n^{2}})$
Therefore, $\displaystyle \frac{a}{b}=\frac{\frac{\sqrt{ab}}{n}(m+\sqrt{m^{2}-n^{2}})}{\frac{\sqrt{ab}}{n}(m-\sqrt{m^{2}-n^{2}})}$
$\displaystyle \frac{a}{b}=\frac{m+\sqrt{m^{2}-n^{2}}}{m-\sqrt{m^{2}-n^{2}}}$
20. If a,b,c are in A.P. b,c,d are in G.P. and $\displaystyle \frac{1}{c},\frac{1}{d},\frac{1}{e}$ are in A.P. prove that a,c,e are in G.P.
Given that, a,b,c are in A.P.
Therefore $\displaystyle b-a=c-b$ ⇒ $\displaystyle 2b=a+c$ ...(i)
And b,c,d are in G.P.
Therefore $\displaystyle c^{2}=bd$ ⇒ $\displaystyle d=\frac{c^{2}}{b}$ ...(ii)
Also $\displaystyle \frac{1}{c},\frac{1}{d},\frac{1}{e}$ in A.P.
Therefore $\displaystyle \frac{1}{d}-\frac{1}{c}=\frac{1}{e}-\frac{1}{d}$
⇒ $\displaystyle \frac{2}{d}=\frac{1}{e}+\frac{1}{c}$ ...(iii)
To prove a,c,e are in G.P. let us prove $\displaystyle c^{2}=ae$
substitute (i) and (ii) in (iii),
$\displaystyle \frac{2}{d}=\frac{1}{e}+\frac{1}{c}$
$\displaystyle \frac{2}{c^{2}/b}=\frac{1}{e}+\frac{1}{c}$
$\displaystyle \frac{2b}{c^{2}}=\frac{c+e}{ce}$
$\displaystyle \frac{a+c}{c}=\frac{c+e}{e}$ [2b=a+c]
$\displaystyle ae+ce=c^{2}+ce$
$\displaystyle c^{2}=ae$
21. Find the sum of the following series upto n terms.
(i) 5+55+555+...
Let $\displaystyle S_{n}=5+55+555+...to \: n\: terms$
$\displaystyle =\frac{5}{9}\left (9+99+999+...to \: n\: terms \right )$
$\displaystyle =\frac{5}{9}\left ((10-1)+(10^{2}-1)+(10^{3}-1)+...to \: n\: terms \right )$
$\displaystyle =\frac{5}{9}\left ((10+10^{2}+10^{3}+...to \: n\: terms)+ (1+1+...to \: n\: terms) \right )$
$\displaystyle =\frac{5}{9}\left ( \frac{10(10^{n}-1)}{10-1}-n \right )$
$\displaystyle =\frac{5}{9}\left ( \frac{10(10^{n}-1)}{9}-n \right )$
$\displaystyle =\frac{50}{81}(10^{n}-1)-\frac{5n}{9}$
(ii) .6+.66+.666+...
Let $\displaystyle S_{n}=.6+.66+.666+...to \: n\: terms$
Let $\displaystyle =6(0.1+0.11+0.111+...to \: n\: terms$
$\displaystyle =\frac{6}{9}\left (0.9+0.99+0.999+...to \: n\: terms \right )$
$\displaystyle =\frac{6}{9}\left [ \left ( 1-\frac{1}{10} \right ) + \left ( 1-\frac{1}{10^{2}} \right )+ \left ( 1-\frac{1}{10^{3}} \right )+to \: n\: terms \right]$
$\displaystyle =\frac{2}{3}\left [ (1+1+to \: n\: terms )-\left ( \frac{1}{10}+\frac{1}{10^{2}}+\frac{1}{10^{3}}+to \: n\: terms \right ) \right]$
$\displaystyle =\frac{2}{3}\left [ n-\frac{1}{10}\left ( 1+\frac{1}{10}+\frac{1}{10^{2}}+to \: n\: terms \right ) \right]$
$\displaystyle =\frac{2}{3}\left [ n-\frac{1}{10}\left ( \frac{1-\frac{1}{10^{n}}}{1-\frac{1}{10}} \right ) \right]$
$\displaystyle =\frac{2}{3}\left [ n-\frac{1}{10}\left ( \frac{1-\frac{1}{10^{n}}}{9/10} \right ) \right]$
$\displaystyle =\frac{2}{3}\left [ n-\left ( \frac{1-10^{-n}}{9} \right ) \right]$
$\displaystyle =\frac{2}{3}n-\frac{2}{27}(1-10^{-n})$
22. Find the $\displaystyle 20^{th}$ term of the series $\displaystyle 2\times 4+4\times 6+6\times 8+..+ n\; terms$
Given the series $\displaystyle 2\times 4+4\times 6+6\times 8+..+ n\; terms$
therefore $\displaystyle n^{th}$ term is $\displaystyle a_{n}=2n\times (2n+2)==4n^{2}+4n$
$\displaystyle a_{20}=4(20)^{2}+4(20)$
$\displaystyle a_{20}=4\times 400+80$
$\displaystyle a_{20}=1600+80=1680$
23. Find the sum of the first n terms of the series: 3+7+13+21+31+...
let $\displaystyle S=3+7+13+21+31+...a_{n-1}+a_{n}$
$\displaystyle S=3+7+13+21+...a_{n-2}+a_{n-1}+a_{n}$
On subtracting both equations,
$\displaystyle S-S=(3+7+13+21+...+a_{n-1}+a_{n})-(3+7+13+21+...a_{n-2}+a_{n-1}+a_{n})$
$\displaystyle 0=3+[(7-3)+(13-7)+(21-13)+...+(a_{n}-a_{n-1})]-a_{n}$
$\displaystyle 0=3+[4+6+8+...+(n-1) terms]-a_{n}$
$\displaystyle a_{n}=3+[4+6+8+...+(n-1) terms]$
$\displaystyle a_{n}=3+\frac{n-1}{2}[2\times 4+(n-1-1)\times 2]$
$\displaystyle =3+\frac{n-1}{2}[8+(n-2)\times 2]$
$\displaystyle =3+\frac{n-1}{2}[8+2n-4]$
$\displaystyle =3+\frac{n-1}{2}[4+2n]$
$\displaystyle =3+(n-1)(n+2)$ $\displaystyle =3+n^{2}+2n-n-2=n^{2}+n+1$
$\displaystyle S_{n}=\sum_{k=1}^{n}a_{k}=\sum_{k=1}^{n}k^{2}+k+1$
$\displaystyle =\sum_{k=1}^{n}k^{2}+\sum_{k=1}^{n}k+\sum_{k=1}^{n}1$
$\displaystyle =\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}+n$
$\displaystyle =n\left [\frac{(n+1)(2n+1)}{6}+\frac{(n+1)}{2}+1 \right ]$
$\displaystyle =n\left [\frac{2n^{2}+2n+1+n}{6}+\frac{(n+1)}{2}+1 \right ]$
$\displaystyle =n\left [\frac{2n^{2}+3n+1+3n+3+6}{6} \right ]$
$\displaystyle =n\left [\frac{2n^{2}+6n+10}{6} \right ]$
$\displaystyle =n\left [\frac{n^{2}+3n+5}{3} \right ]$
24. If $\displaystyle S_{1},S_{2},S_{3}$ are the sum of first n natural numbers, their squares and their cubes respectively, show that $\displaystyle 9S_{2}^{2}=S_{3}(1+8S_{1})$
Given that,
$\displaystyle S_{1}=\frac{n(n+1)}{2}$
$\displaystyle S_{2}=\frac{n(n+1)(2n+1)}{6}$
$\displaystyle S_{3}=\frac{n^{2}(n+1)^{2}}{4}$
To prove $\displaystyle 9S_{2}^{2}=S_{3}(1+8S_{1})$,
consider L.H.S.
$\displaystyle 9S_{2}=9\left [\frac{n(n+1)(2n+1)}{6} \right ]^{2}$
$\displaystyle 9S_{2}=9\left [\frac{n^{2}(n+1)^{2}(2n+1)^{2}}{36} \right ]$
$\displaystyle 9S_{2}=\frac{n^{2}(n+1)^{2}(2n+1)^{2}}{4}$ ...(i)
consider R.H.S.
$\displaystyle S_{3}(1+8S_{1})=\frac{n^{2}(n+1)^{2}}{4}\left [ 1+8\left ( \frac{n(n+1)}{2} \right ) \right ]$
$\displaystyle =\frac{n^{2}(n+1)^{2}}{4}\left [ 1+4n(n+1) \right ]$
$\displaystyle =\frac{n^{2}(n+1)^{2}}{4}\left [ 1+4n^{2}+4n \right ]$
$\displaystyle =\frac{n^{2}(n+1)^{2}}{4}\left [ (2n+1)^{2} \right ]$
$\displaystyle S_{3}(1+8S_{1})=\frac{n^{2}(n+1)^{2}(2n+1)^{2} }{4}$ ...(ii)
from (i) and (ii), L.H.S.=R.H.S.
25. Find the sum of the following series up to n terms $\displaystyle \frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+...$
From the series $\displaystyle n^{th}$ term is $\displaystyle \frac{1^{3}+2^{3}+3^{3}+...+n^{3}}{1+3+5+...+(2n-1)}$
$\displaystyle =\frac{\left ( \frac{n^{2}(n+1)^{2}}{4} \right )}{n^{2}}$
$\displaystyle = \frac{n^{2}(n+1)^{2}}{4n^{2}}=\frac{(n+1)^{2}}{4}$
$\displaystyle =\frac{n^{2}+1+2n}{4}$
$\displaystyle =\frac{1}{4}n^{2}+\frac{1}{2}n+\frac{1}{4}$
$\displaystyle S_{n}=\sum_{k=1}^{n}a_{k}=\sum_{k=1}^{n}(\frac{1}{4}k^{2}+\frac{1}{2}k+\frac{1}{4})$
$\displaystyle =\frac{1}{4}\sum_{k=1}^{n}k^{2}+\frac{1}{2}\sum_{k=1}^{n}k+\frac{1}{4}\sum_{k=1}^{n}1$
$\displaystyle =\frac{1}{4}\left [ \frac{n(n+1)(2n+1)}{6} \right ]+\frac{1}{2}\left [ \frac{n(n+1)}{2} \right ]+\frac{1}{4}n$
$\displaystyle =\frac{n(n+1)(2n+1)}{24} + \frac{n(n+1)}{4} +\frac{n}{4}$
$\displaystyle =\frac{n[(n+1)(2n+1)+6(n+1)+6]}{24}$
$\displaystyle =\frac{n[2n^{2}+n+2n+1+6n+6+6]}{24}$
$\displaystyle =\frac{n[2n^{2}+9n+13]}{24}$
26. Show that $\displaystyle \frac{1\times 2^{2}+2\times 3^{2}+...n\times (n+1)^{2}}{1^{2}\times 2+2^{2}\times 3+...+n^{2}\times (n+1)}=\frac{3n+5}{3n+1}$
consider numerator, $\displaystyle 1\times 2^{2}+2\times 3^{2}+...n\times (n+1)^{2}$
$\displaystyle n^{th}$ term is $\displaystyle n(n+1)^{2}=n(n^{2}+1+2n)=n^{3}+2n^{2}+n$
$\displaystyle S_{n}=\sum_{k=1}^{n}a_{k}=\sum_{k=1}^{n}(k^{3}+2k^{2}+k)$
$\displaystyle =\sum_{k=1}^{n}k^{3}+2\sum_{k=1}^{n}k^{2}+\sum_{k=1}^{n}k$
$\displaystyle =\frac{n^{2}(n+1)^{2}}{4}+2\left ( \frac{n(n+1)(2n+1)}{6} \right )+\frac{n(n+1)}{2}$
$\displaystyle =\frac{n^{2}(n+1)^{2}}{4}+ \frac{2n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}$
$\displaystyle =\frac{n(n+1)}{2}\left [\frac{n(n+1)}{2}+ \frac{2(2n+1)}{3}+1 \right ]$
$\displaystyle =\frac{n(n+1)}{2}\left [\frac{n^{2}+n}{2}+ \frac{4n+2}{3}+1 \right ]$
$\displaystyle =\frac{n(n+1)}{2}\left [\frac{3n^{2}+3n+8n+4+6}{6} \right ]$
$\displaystyle =\frac{n(n+1)}{2}\left [\frac{3n^{2}+11n+10}{6} \right ]$
$\displaystyle =\frac{n(n+1)}{12}[(n+2)(3n+5)]$
$\displaystyle =\frac{n(n+1)(n+2)(3n+5)}{12}$ ...(i)
consider denominator, $\displaystyle 1^{2}\times 2+2^{2}\times 3+...+n^{2}\times (n+1)$
$\displaystyle n^{th}$ term is $\displaystyle n^{2}(n+1)=n^{3}+n^{2}$
$\displaystyle S_{n}=\sum_{k=1}^{n}a_{k}=\sum_{k=1}^{n}(k^{3}+k^{2})$
$\displaystyle =\sum_{k=1}^{n}k^{3}+\sum_{k=1}^{n}k^{2}$
$\displaystyle =\frac{n^{2}(n+1)^{2}}{4}+\frac{n(n+1)(2n+1)}{6}$
$\displaystyle =\frac{n(n+1)}{2}\left [\frac{n(n+1)}{2}+ \frac{2n+}{3} \right ]$
$\displaystyle =\frac{n(n+1)}{2}\left [\frac{n^{2}+n}{2}+ \frac{2n+}{3} \right ]$
$\displaystyle =\frac{n(n+1)}{2}\left [\frac{3n^{2}+3n+4n+2}{6} \right ]$
$\displaystyle =\frac{n(n+1)}{2}\left [\frac{3n^{2}+7n+2}{6} \right ]$
$\displaystyle =\frac{n(n+1)}{12}(3n^{2}+7n+2)$
$\displaystyle =\frac{n(n+1)}{12}(n+2)(3n+1)$
$\displaystyle =\frac{n(n+1)(n+2)(3n+1)}{12}$ ...(ii)
Using (i) and (ii),
$\displaystyle \frac{1\times 2^{2}+2\times 3^{2}+...n\times (n+1)^{2}}{1^{2}\times 2+2^{2}\times 3+...+n^{2}\times (n+1)}$
$\displaystyle =\frac{\frac{n(n+1)(n+2)(3n+5)}{12}}{\frac{n(n+1)(n+2)(3n+1)}{12}}$
$\displaystyle =\frac{(3n+5)}{(3n+1)}$
27. A farmer buys a used tractor for Rs. 12000. He pays Rs. 6000 cash and agrees to pay the balance in annual instalments of Rs. 500 plus 12% interest on the unpaid amount. How much will the tractor cost him?
Given, cost of tractor = 12000
amount paid = 6000
amount unpaid = 12000-6000 = 6000
Interest paid annually is,
12% of 6000, 12% of 5500, 12% of 5000,...12% of 500
Total interest to be paid is
=12% of 6000+12% of 5500+12% of 5000+...+12% of 500
=12%[6000+5500+5000+...+500]
=12%[500+1000+...+6000] ...(i)
consider the series 500,1000,...,6000
let the number of terms be n and a=500, l=6000, d=500
We know that,
l=a+(n-1)d
6000=500+(n-1)500
6000=500[1+n-1]
n=12
$\displaystyle S_{n}=\frac{n}{2}(2a+(n-1)d)$
$\displaystyle =\frac{12}{2}(2(500)+(12-1)500)$
$\displaystyle =6(1000+11 \times 500)$
$\displaystyle =6(1000+5500)=39000$
Interest to be paid is 12%[500+1000+...+6000]=12% (39000)=4680
Total cost of tractor = 12000+4680=16680
28. Shamshad Ali buys a scooter for Rs. 22000. He pays Rs. 4000 cash and agrees to pay the balance in annual instalment of Rs. 1000 plus 10% interest on the unpaid amount. How much will the tractor cost him?
Given, cost of scooter = 22000
amount paid = 4000
amount unpaid = 22000-4000 = 18000
Interest paid annually is,
10% of 18000, 10% of 17000, 10% of 16000,...10% of 1000
Total interest to be paid is
=10% of 18000+10% of 17000+10% of 16000,...+10% of 1000
=10%[18000+17000+...+1000]
=10%[1000+2000+...+18000] ...(i)
consider the series 1000,2000,...,18000
let the number of terms be n and a=1000, l=18000, d=1000
We know that,
l=a+(n-1)d
18000=1000+(n-1)1000
18000=1000[1+n-1]
n=18
$\displaystyle S_{n}=\frac{n}{2}(2a+(n-1)d)$
$\displaystyle =\frac{18}{2}(2(1000)+(18-1)1000)$
$\displaystyle =9(2000+17 \times 1000)$
$\displaystyle =9(2000+17000)=171000$
Interest to be paid is 10%[1000+2000+...+18000]=10% (171000)=17100
Total cost of tractor = 22000+17100=39100
29. A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8th set of letter is mailed.
Number of letters mailed is $\displaystyle 4,4^{2},4^{3},...4^{8}$
a=4, r=4, n=8
$\displaystyle S_{n}=\frac{a(r^{n}-1)}{r-1}$
$\displaystyle =\frac{4(4^{8}-1)}{4-1}$
$\displaystyle =\frac{4(65536-1)}{3}$
$\displaystyle =\frac{4(65535)}{3}=87380$
Given the cost of one mail is 50paise,
Cost of 87380 is $\displaystyle =87380\frac{50}{100}=43690$
30. A man deposited Rs. 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years.
Amount deposited = 10000
rate of interest = 5%
interest amount = 5% of 10000 = 500 Rs
Amount in 15th year is $\displaystyle =10000+14 \times 500 = 17000$
Amount after 20 year is $\displaystyle =10000+20 \times 500 = 20000$
31. A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.
Cost of machine = 15625 Rs
Machine value depreciates by 20% every year.
therefore, its value after every year is 80% less than originial cost
⇒ $\displaystyle \frac{80}{100}=\frac{4}{5}$
Value at the end of 5 years is
$\displaystyle =5\times \frac{4}{5}\times 15625=5120$
32. 150 workers were engaged to finish a job in a certain number of days, 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.
Let x be the number of days in which 150 workers finish the work.
Given that,
150x=150+146+142+.. (x+8) terms
the series 150+146+142... is an A.P.
a=150, d=-4, n=x+8
$\displaystyle 150x=\frac{x+8}{2}[2(150)+(x+8-1)(-4)]$
$\displaystyle 150x=(x+8)[150+(x+7)(-2)]$
$\displaystyle 75x=(x+8)[75+(x+7)(-1)]$
$\displaystyle 75x=(x+8)[75-x-7]$
$\displaystyle 75x=(x+8)[68-x]$
$\displaystyle 75x=68x-x^{2}+544-8x$
$\displaystyle x^{2}-135x-544=0$
$\displaystyle (x-17)(x+32)=0$
$\displaystyle x=17$ or $\displaystyle x=-32$
x cannot be negative therefore $\displaystyle x=17$
The number of days required= 17+8=25
Let a be the first term and d be the common difference of A.P.
We know that $\displaystyle k^{th}$ term of an A.P. is $\displaystyle a_{k}=a+(k-1)d$
⇒ $\displaystyle a_{m+n}=a+(m+n-1)d$ and $\displaystyle a_{m-n}=a+(m-n-1)d$
$\displaystyle a_{m+n}+a_{m-n}$ $\displaystyle =a+(m+n-1)d+a$ $\displaystyle +(m-n-1)d$
$\displaystyle =2a+(m+n-1+m-n-1)d$
$\displaystyle =2a+(2m-2)d$
$\displaystyle =2[a+(m-1)d]$
$\displaystyle =2a_{m}$
Thus sum of $\displaystyle (m+n)^{th}$ and $\displaystyle (m-n)^{th}$ terms is equal to twice the $\displaystyle m^{th}$ term.
2. If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.
Let the three numbers be a-d, a and a+d
Given that,
$\displaystyle (a-d)+a+(a+d)=24$
$\displaystyle 3a=24$
$\displaystyle a=8$
Also given that,
$\displaystyle (a-d)a(a+d)=440$
since a=8,
$\displaystyle (8-d)8(8+d)=440$
$\displaystyle (8-d)(8+d)=55$
$\displaystyle 8^{2}-d^{2}=55$
$\displaystyle 64-d^{2}=55$
$\displaystyle d^{2}=64-55=9$
$\displaystyle d=\pm 3$
When d=3, the numbers are 5, 8, 11 and when d=-3 the numbers are 11, 8, 5.
Hence, the numbers are 5, 8 and 11
3. Let the sum of n, 2n, 3n terms of an A.P. be $\displaystyle S_{1},S_{2},S_{3}$ respectively, show that $\displaystyle S_{3}=3(S_{2}-S_{1})$
Let a be the first term and d be the common difference of A.P.
$\displaystyle S_{1}=\frac{n}{2}(2a+(n-1)d)$
$\displaystyle S_{2}=\frac{2n}{2}(2a+(2n-1)d)$
$\displaystyle S_{1}=\frac{3n}{2}(2a+(3n-1)d)$
Consider $\displaystyle 3(S_{2}-S_{1})$
$\displaystyle =3\left [ \frac{2n}{2}(2a+(2n-1)d) -\frac{n}{2}(2a+(n-1)d)\right ]$
$\displaystyle =\frac{3n}{2}\left [ 2(2a+(2n-1)d) -(2a+(n-1)d)\right ]$
$\displaystyle =\frac{3n}{2}\left [ (4a+4nd-2d) -(2a+nd-d)\right ]$
$\displaystyle =\frac{3n}{2}\left [ (4a+4nd-2d-2a-nd+d)\right ]$
$\displaystyle =\frac{3n}{2}(2a+3nd-d)$
$\displaystyle =\frac{3n}{2}(2a+(3n-1)d)=S_{3}$
4. Find the sum of all numbers between 200 and 400 which are divisible by 7.
The numbers between 200 and 400 which are divisible by 7 are 203, 210,...,399.
Therefore, a=203
l=399
d=7
Let the numbers of terms be n
$\displaystyle a_{n}=a+(n-1)d=399$
$\displaystyle 203+(n-1)7=399$
$\displaystyle (n-1)7=399-203=196$
$\displaystyle (n-1)=\frac{196}{7}=28$
$\displaystyle n=29$
$\displaystyle S_{n}=\frac{n}{2}(a+l)$
$\displaystyle =\frac{29}{2}(203+399)$
$\displaystyle =\frac{29}{2}(602)=8729$
5. Find the sum of integers from 1 to 100 that are divisible by 2 or 5.
The integers from 1 to 100 that are divisible by 2 are 2,4,..100.
Therefore, a=2, l=100, d=2
$\displaystyle a_{n}=a+(n-1)d$
$\displaystyle 100=2+(n-1)d$
$\displaystyle 100-2=(n-1)2$
$\displaystyle 98=(n-1)2$
$\displaystyle n-1=49$ ⇒ $\displaystyle n=50$
$\displaystyle S_{n}=\frac{n}{2}(a+l)$
$\displaystyle =\frac{50}{2}(2+100)$
$\displaystyle =25(102)=2550$
The integers from 1 to 100 that are divisible by 5 are 5,10,..100.
Therefore, a=5, l=100, d=5
$\displaystyle a_{n}=a+(n-1)d$
$\displaystyle 100=5+(n-1)5$
$\displaystyle 100-5=(n-1)5$
$\displaystyle 95=(n-1)5$
$\displaystyle n-1=19$ ⇒ $\displaystyle n=20$
$\displaystyle S_{n}=\frac{n}{2}(a+l)$
$\displaystyle =\frac{20}{2}(5+100)$
$\displaystyle =10(105)=1050$
The integers from 1 to 100 that are divisible by 2 and 5 are 10,20,..100.
Therefore, a=10, l=100, d=10
$\displaystyle a_{n}=a+(n-1)d$
$\displaystyle 100=10+(n-1)10$
$\displaystyle 100-10=(n-1)10$
$\displaystyle 90=(n-1)10$
$\displaystyle n-1=9$ ⇒ $\displaystyle n=10$
$\displaystyle S_{n}=\frac{n}{2}(a+l)$
$\displaystyle =\frac{10}{2}(10+100)$
$\displaystyle =5(110)=550$
Therefore the required sum is $\displaystyle 2550+1050-550=3050$
6. Find the sum of all two digit numbers which when divided by 4 yields 1 as remainder.
Two digit numbers which when divided by 4 yields 1 as remainder are 13,17,...97
Therefore, a=13, l=97, d=4
$\displaystyle a_{n}=a+(n-1)d$
$\displaystyle 97=13+(n-1)4$
$\displaystyle 97-13=(n-1)4$
$\displaystyle 84=(n-1)4$
$\displaystyle n-1=21$ ⇒ $\displaystyle n=22$
$\displaystyle S_{n}=\frac{n}{2}(a+l)$
$\displaystyle =\frac{22}{2}(13+97)$
$\displaystyle =11(110)=1210$
7. If f is a function satisfying $\displaystyle f(x+y)=f(x)f(y)$ for all $\displaystyle x,y\epsilon N$ such that f(1)=3 and $\displaystyle \sum_{x=1}^{n}f(x)=120$, find the value of n.
Given that, $\displaystyle f(x+y)=f(x)f(y)$ and f(1)=3
Taking x=y=1,
$\displaystyle f(1+1)=f(1)f(1)$
$\displaystyle f(2)=3\times 3=3^{2}=9$
Similarly,
$\displaystyle f(3)=3^{3}=27$
$\displaystyle f(4)=3^{4}=81$
f(1), f(2), f(3),... i.e. 3,9,27,... forms a G.P.
⇒ a=3, r=3
$\displaystyle \sum_{x=1}^{n}f(x)=120$
$\displaystyle f(1)+f(2)+f(3)+...+f(n)=120$
$\displaystyle 3+3^{2}+3^{3}+...+3^{n}=120$
$\displaystyle \frac{3(3^{n}-1)}{3-1}=120$
$\displaystyle \frac{3(3^{n}-1)}{2}=120$
$\displaystyle 3(3^{n}-1)=240$
$\displaystyle 3^{n}-1=80$
$\displaystyle 3^{n}=81$
⇒ $\displaystyle 3^{n}=3^{4}$
⇒ n=4
8. The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2 respectively. Find the last term and the number of terms.
Given that, a=5, r=2
Let sum of n terms be 315.
$\displaystyle S_{n}=\frac{a(r^{n}-1)}{r-1}=315$
$\displaystyle \frac{5(2^{n}-1)}{2-1}=315$
$\displaystyle 5(2^{n}-1)=315$
$\displaystyle 2^{n}-1=63$
$\displaystyle 2^{n}=64=2^{6}$
Therefore, number of terms n=6
Last term = $\displaystyle ar^{n-1}$
$\displaystyle 5(2)^{6-1}$
$\displaystyle 5(2)^{5}=5\times 32=160$
9. The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.
Given that, a=1,
$\displaystyle a_{3}+a_{5}=90$
$\displaystyle ar^{2}+ar^{4}=90$
$\displaystyle r^{2}+r^{4}=90$
$\displaystyle r^{4}+r^{2}-90=0$
$\displaystyle r^{2}=\frac{1\pm \sqrt{1+4\times 90}}{2}$
$\displaystyle r^{2}=\frac{1\pm \sqrt{1+360}}{2}$
$\displaystyle r^{2}=\frac{1\pm \sqrt{361}}{2}$ $\displaystyle =\frac{1\pm 19}{2}$ $\displaystyle =-10 or 9$
$\displaystyle r=\pm 3$
10. The sum of three numbers in G.P. is 56. If we subtract 1,7,21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.
Let the numbers be $\displaystyle a,ar,ar^{2}$
Given that, $\displaystyle a+ar+ar^{2}=56$ ...(i)
and $\displaystyle a-1, ar-7,ar^{2}-21$ form an A.P.
⇒ $\displaystyle (ar-7)-(a-1)=(ar^{2}-21)-(ar-7)$
$\displaystyle ar-a-6=ar^{2}-ar-14$
$\displaystyle ar^{2}-2ar+a=8$
$\displaystyle a(r^{2}-2r+1)=8$ ...(ii)
$\displaystyle \frac{(i)}{(ii)}=\frac{a(1+r+r^{2})}{a(r^{2}-2r+1)}=\frac{56}{8}$
$\displaystyle \frac{1+r+r^{2}}{r^{2}-2r+1}=7$
$\displaystyle 1+r+r^{2}=7(r^{2}-2r+1)$
$\displaystyle 6r^{2}-15r+6=0$
$\displaystyle 2r^{2}-5r+2=0$
$\displaystyle (r-2)(2r-1)=0$
⇒ $\displaystyle r=2\: or\: \frac{1}{2}$
when $\displaystyle r=2$ , (i) ⇒ a=8 and three numbers are 8,16,32
when $\displaystyle r=\frac{1}{2}$ , (i) ⇒ a=32 and three numbers are 32,16,8
11. A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.
Let the number of terms be 2n ⇒ G.P. is $\displaystyle G_{1},G_{2},G_{3}...G_{2n}$
Given that, $\displaystyle G_{1}+G_{2}+G_{3}...+G_{2n}=5[G_{1}+G_{3}...+G_{2n-1}]$
$\displaystyle G_{1}+G_{2}+G_{3}...+G_{2n}-5[G_{1}+G_{3}...+G_{2n-1}]=0$
$\displaystyle G_{2}+G_{4}...+G_{2n}-4[G_{1}+G_{3}...+G_{2n-1}]=0$
$\displaystyle G_{2}+G_{4}...+G_{2n}=4[G_{1}+G_{3}...+G_{2n-1}]$
$\displaystyle ar+ar^{3}+...ar^{2n-1}=4[a+ar^{2}+...ar^{2n-2}]$
⇒ r=4
12. The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.
Let the first four terms are a,a+d,a+2d,a+3d
and the last four terms are a+(n-4)d,a+(n-3)d,a+(n-2)d,a+(n-1)d
Given that,
$\displaystyle a+a+d+a+2d+a+3d=56$
$\displaystyle 4a+6d=56$
$\displaystyle 4(11)+6d=56$
$\displaystyle 44+6d=56$
$\displaystyle 6d=12$ ⇒ $\displaystyle d=2$
Also given that,
$\displaystyle a+(n-4)d+a+(n-3)d+a+(n-2)d+a+(n-1)d=112$
$\displaystyle 4a+(4n-10)d=112$
$\displaystyle 4(11)+(4n-10)2=112$
$\displaystyle 44+8n-20=112$
$\displaystyle 8n=88$ ⇒ $\displaystyle n=11$
13. If $\displaystyle \frac{a+bx}{a-bx}=\frac{b+cx}{b-cx}=\frac{c+dx}{c-dx}$ (xne0), then show that a,b,c and d are in G.P.
Given that,
$\displaystyle \frac{a+bx}{a-bx}=\frac{b+cx}{b-cx}$
$\displaystyle (a+bx)(b+cx)=(b-cx)(a-bx)$
$\displaystyle ab-acx+b^{2}x-bcx^{2}=ab-b^{2}x+acx-bcx^{2}$
$\displaystyle 2b^{2}x=2acx$
⇒ $\displaystyle \frac{b}{c}=\frac{a}{b}$ ...(i)
Also given that,
$\displaystyle \frac{b+cx}{b-cx}=\frac{c+dx}{c-dx}$
$\displaystyle (b+cx)(c+dx)=(c-dx)(b-cx)$
$\displaystyle bc-bdx+c^{2}x-cdx^{2}=bc-c^{2}x+bdx-cdx^{2}$
$\displaystyle 2c^{2}x=2bdx$
⇒ $\displaystyle \frac{c}{d}=\frac{b}{c}$ ...(ii)
from (i) and (ii), $\displaystyle \frac{a}{b}=\frac{b}{c}=\frac{c}{d}$
⇒ a,b,c and are in G.P.
14. Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that $\displaystyle P^{2}R^{n}=S^{n}$
Let the G.P. be $\displaystyle a,ar,ar^{2},...ar^{n-1}$
Given that,
$\displaystyle S=\frac{a(r^{n}-1)}{r-1}$
$\displaystyle P=a^{n}r^{1+2+...n-1}=a^{n}r^{\frac{n(n-1)}{2}}$
$\displaystyle R=\frac{1}{a}+\frac{1}{ar}+...\frac{1}{ar^{n-1}}$
$\displaystyle R=\frac{r^{n-1}+r^{n-2}+...r+1}{ar^{n-1}}$
$\displaystyle =(1+r+...r^{n-2}+r^{n-1})\frac{1}{ar^{n-1}}$
$\displaystyle =\frac{1(r^{n}-1)}{r-1}\times \frac{1}{ar^{n-1}}$
$\displaystyle =\frac{r^{n}-1}{ar^{n-1}(r-1)}$
Therefore,
$\displaystyle P^{2}R^{n}=a^{2n}r^{\frac{2n(n-1)}{2}}\times \frac{(r^{n}-1)^{n}}{a^{n}r^{n(n-1)}(r-1)^{n}}$
$\displaystyle =\frac{a^{2n}r^{n(n-1)}(r^{n}-1)^{n}}{a^{n}r^{n(n-1)}(r-1)^{n}}$
$\displaystyle =\frac{a^{n}(r^{n}-1)^{n}}{(r-1)^{n}}$
$\displaystyle =\left [\frac{a(r^{n}-1)}{r-1} \right ]^{n}=S^{n}$
15. The $\displaystyle p^{th},q^{th},r^{th}$ terms of an A.P. are a,b,c respectively. Show that $\displaystyle (q-r)a+(r-p)b+(p-q)c=0$
Let f be the first term and d be the common difference.
Given that,
$\displaystyle a_{p}=f+(p-1)d=a$ ...(i)
$\displaystyle a_{q}=f+(q-1)d=b$ ...(ii)
$\displaystyle a_{r}=f+(r-1)d=c$ ...(iii)
(i)-(ii) ⇒
$\displaystyle a-b=f+(p-1)d-(f+(q-1)d)$
$\displaystyle a-b=(p-1)d-(q-1)d$
$\displaystyle a-b=(p-1-q+1)d$
$\displaystyle a-b=(p-q)d$ ...(iv)
(ii)-(iii) ⇒
$\displaystyle b-c=f+(q-1)d-(f+(r-1)d)$
$\displaystyle b-c=(q-1)d-(r-1)d$
$\displaystyle b-c=(q-1-r+1)d$
$\displaystyle b-c=(q-r)d$ ...(v)
Equatin (iv)and (v) ⇒
$\displaystyle \frac{a-b}{p-q}=\frac{b-c}{q-r}$
$\displaystyle (a-b)(q-r)=(b-c)(p-q)$
$\displaystyle aq-ar-bq+br=bp-bq-cp+cq$
$\displaystyle aq-ar-bq+br-bp+bq+cp-cq=0$
$\displaystyle a(q-r)+b(-q+r-p+q)+c(p-q)=0$
$\displaystyle a(q-r)+b(r-p)+c(p-q)=0$
16. If $\displaystyle a\left ( \frac{1}{b}+\frac{1}{c} \right )$, $\displaystyle b\left ( \frac{1}{c}+\frac{1}{a} \right )$, $\displaystyle c\left ( \frac{1}{a}+\frac{1}{b} \right )$ are in A.P. prove that a,b,c are in A.P.
Given that,
$\displaystyle a\left ( \frac{1}{b}+\frac{1}{c} \right )$, $\displaystyle b\left ( \frac{1}{c}+\frac{1}{a} \right )$, $\displaystyle c\left ( \frac{1}{a}+\frac{1}{b} \right )$ are in A.P. ⇒
$\displaystyle b\left ( \frac{1}{c}+\frac{1}{a} \right )$-$\displaystyle a\left ( \frac{1}{b}+\frac{1}{c} \right )$=$\displaystyle c\left ( \frac{1}{a}+\frac{1}{b} \right )$-$\displaystyle b\left ( \frac{1}{c}+\frac{1}{a} \right )$
$\displaystyle b\left ( \frac{a+c}{ac} \right )-a\left ( \frac{b+c}{bc} \right )=c\left ( \frac{a+b}{ab} \right )-b\left ( \frac{a+c}{ac} \right )$
$\displaystyle \left ( \frac{ab+bc}{ac} \right )-\left ( \frac{ab+ac}{bc} \right )=\left ( \frac{ac+bc}{ab} \right )-\left ( \frac{ab+bc}{ac} \right )$
$\displaystyle \frac{ab^{2}+b^{2}c-a^{2}b-a^{2}c}{abc} = \frac{ac^{2}+bc^{2}-ab^{2}-b^{2}c}{abc}$
$\displaystyle ab^{2}+b^{2}c-a^{2}b-a^{2}c=ac^{2}+bc^{2}-ab^{2}-b^{2}c$
$\displaystyle ab(b-a)+c(b^{2}-a^{2})=a(c^{2}-b^{2})+bc(c-b)$
$\displaystyle ab(b-a)+c(b-a)(b+a)=a(c-b)(c+b)+bc(c-b)$
$\displaystyle (b-a)(ab+cb+ca)=(c-b)(ac+ab+bc)$
$\displaystyle b-a=c-b$
Thus a,b,c are in A.P.
17. If a,b,c,d are in G.P. prove that $\displaystyle (a^{n}+b^{n}),(b^{n}+c^{n}),(c^{n}+d^{n})$ are in G.P.
Given that a,b,c,d are in G.P.
$\displaystyle \frac{c}{b}=\frac{b}{a}$ ⇒ $\displaystyle b^{2}=ac$ ...(i)
$\displaystyle \frac{d}{c}=\frac{c}{b}$ ⇒ $\displaystyle c^{2}=bd$ ...(ii)
$\displaystyle \frac{b}{a}=\frac{d}{c}$ ⇒ $\displaystyle bc=ad$ ...(iii)
To prove $\displaystyle (a^{n}+b^{n}),(b^{n}+c^{n}),(c^{n}+d^{n})$ are in G.P., let us prove $\displaystyle (b^{n}+c^{n})^{2}=(a^{n}+b^{n})(c^{n}+d^{n})$
Consider L.H.S.
$\displaystyle (b^{n}+c^{n})^{2}=b^{2n}+c^{2n}+2b^{n}c^{n}$
$\displaystyle =(b^{2})^{n}+(c^{2})^{n}+2(bc)^{n}$
$\displaystyle =(b^{2})^{n}+(c^{2})^{n}+(bc)^{n}+(bc)^{n}$
$\displaystyle =(ac)^{n}+(bd)^{n}+(bc)^{n}+(ad)^{n}$ [using (i),(ii),(iii)]
$\displaystyle =a^{n}c^{n}+b^{n}c^{n}+b^{n}d^{n}+a^{n}d^{n}$
$\displaystyle =c^{n}(a^{n}+b^{n})+d^{n}(a^{n}+b^{n})$
$\displaystyle =(a^{n}+b^{n})(c^{n}+d^{n})$
L.H.S.=R.H.S.
18. If a and b are the roots of $\displaystyle x^{2}-3x+p=0$ and c,d are roots of $\displaystyle x^{2}-12x+q=0$ where a,b,c,d form a G.P. Prove that $\displaystyle (q+p):(q-p)=17:15$
Given that, a,b are roots of $\displaystyle x^{2}-3x+p=0$
therefore a+b=3 and ab=p ...(i)
And c,d are roots of $\displaystyle x^{2}-12x+q=0$
therefore c+d=12 and cd=q ...(ii)
Also given that a,b,c,d are in G.P.
Let $\displaystyle a=x,b=xr,c=xr^{2},d=xr^{3}$
From (i) and (ii)
$\displaystyle x+xr=3$ ⇒ $\displaystyle x(1+r)=3$ ...(iii)
$\displaystyle xr^{2}+xr^{3}=12$ ⇒ $\displaystyle xr^{2}(1+r)=12$ ...(iv)
Dividing (iv) by (iii),
$\displaystyle \frac{xr^{2}(1+r)}{x(1+r)}=\frac{12}{3}$
$\displaystyle r^{2}=4$ ⇒ $\displaystyle r=\pm 2$
Case I: when $\displaystyle r=2, x=\frac{3}{1+2}=1$
from (i) and (ii)
$\displaystyle p=ab=x^{2}r=2$
$\displaystyle q=cd=x^{2}r^{5}=32$
$\displaystyle \frac{q+p}{q-p}=\frac{32+2}{32-2}=\frac{34}{30}=\frac{17}{15}$
Case II: when $\displaystyle r=-2, x=\frac{3}{1-2}=-3$
from (i) and (ii)
$\displaystyle p=ab=x^{2}r=-18$
$\displaystyle q=cd=x^{2}r^{5}=-288$
$\displaystyle \frac{q+p}{q-p}=\frac{-288-18}{-288+18}=\frac{-306}{-270}=\frac{17}{15}$
19. The ratio of the A.M. and G.M. of two positive numbers a and b is m:n. Show that $\displaystyle a:b=(m+\sqrt{m^{2}-n^{2}}):(m-\sqrt{m^{2}-n^{2}})$
Consider the numbers a,b
A.M.=$\displaystyle \frac{a+b}{2}$ and G.M.=$\displaystyle \sqrt{ab}$
Given that, $\displaystyle \frac{AM}{GM}=\frac{m}{n}$
$\displaystyle \frac{a+b}{2\sqrt{ab}}=\frac{m}{n}$
$\displaystyle a+b=\frac{2\sqrt{ab}m}{n}$ ...(i)
Consider,
$\displaystyle (a-b)^{2}=(a+b)^{2}-4ab$
$\displaystyle (a-b)^{2}=\left ( \frac{2\sqrt{ab}m}{n} \right )^{2}-4ab$
$\displaystyle (a-b)^{2}=\frac{4abm^{2}}{n^{2}} -4ab$
$\displaystyle (a-b)^{2}=4ab\left ( \frac{m^{2}}{n^{2}} -1 \right )$
$\displaystyle (a-b)^{2}=4ab\left ( \frac{m^{2}-n^{2}}{n^{2}} \right )$
$\displaystyle a-b=2\sqrt{ab}\left ( \frac{\sqrt{m^{2}-n^{2}}}{n} \right )$ ...(ii)
Adding (i) and (ii)
$\displaystyle a+b+a-b=\frac{2\sqrt{ab}m}{n}+2\sqrt{ab}\left ( \frac{\sqrt{m^{2}-n^{2}}}{n} \right )$
$\displaystyle 2a=\frac{2\sqrt{ab}}{n}(m+\sqrt{m^{2}-n^{2}})$
$\displaystyle a=\frac{\sqrt{ab}}{n}(m+\sqrt{m^{2}-n^{2}})$
Substitute value of a in (i) ⇒
$\displaystyle b=\frac{2\sqrt{ab}m}{n}-a$
$\displaystyle b=\frac{2\sqrt{ab}m}{n}-\frac{\sqrt{ab}}{n}(m+\sqrt{m^{2}-n^{2}})$
$\displaystyle b=\frac{\sqrt{ab}}{n}(2m-m-\sqrt{m^{2}-n^{2}})$
$\displaystyle b=\frac{\sqrt{ab}}{n}(m-\sqrt{m^{2}-n^{2}})$
Therefore, $\displaystyle \frac{a}{b}=\frac{\frac{\sqrt{ab}}{n}(m+\sqrt{m^{2}-n^{2}})}{\frac{\sqrt{ab}}{n}(m-\sqrt{m^{2}-n^{2}})}$
$\displaystyle \frac{a}{b}=\frac{m+\sqrt{m^{2}-n^{2}}}{m-\sqrt{m^{2}-n^{2}}}$
20. If a,b,c are in A.P. b,c,d are in G.P. and $\displaystyle \frac{1}{c},\frac{1}{d},\frac{1}{e}$ are in A.P. prove that a,c,e are in G.P.
Given that, a,b,c are in A.P.
Therefore $\displaystyle b-a=c-b$ ⇒ $\displaystyle 2b=a+c$ ...(i)
And b,c,d are in G.P.
Therefore $\displaystyle c^{2}=bd$ ⇒ $\displaystyle d=\frac{c^{2}}{b}$ ...(ii)
Also $\displaystyle \frac{1}{c},\frac{1}{d},\frac{1}{e}$ in A.P.
Therefore $\displaystyle \frac{1}{d}-\frac{1}{c}=\frac{1}{e}-\frac{1}{d}$
⇒ $\displaystyle \frac{2}{d}=\frac{1}{e}+\frac{1}{c}$ ...(iii)
To prove a,c,e are in G.P. let us prove $\displaystyle c^{2}=ae$
substitute (i) and (ii) in (iii),
$\displaystyle \frac{2}{d}=\frac{1}{e}+\frac{1}{c}$
$\displaystyle \frac{2}{c^{2}/b}=\frac{1}{e}+\frac{1}{c}$
$\displaystyle \frac{2b}{c^{2}}=\frac{c+e}{ce}$
$\displaystyle \frac{a+c}{c}=\frac{c+e}{e}$ [2b=a+c]
$\displaystyle ae+ce=c^{2}+ce$
$\displaystyle c^{2}=ae$
21. Find the sum of the following series upto n terms.
(i) 5+55+555+...
Let $\displaystyle S_{n}=5+55+555+...to \: n\: terms$
$\displaystyle =\frac{5}{9}\left (9+99+999+...to \: n\: terms \right )$
$\displaystyle =\frac{5}{9}\left ((10-1)+(10^{2}-1)+(10^{3}-1)+...to \: n\: terms \right )$
$\displaystyle =\frac{5}{9}\left ((10+10^{2}+10^{3}+...to \: n\: terms)+ (1+1+...to \: n\: terms) \right )$
$\displaystyle =\frac{5}{9}\left ( \frac{10(10^{n}-1)}{10-1}-n \right )$
$\displaystyle =\frac{5}{9}\left ( \frac{10(10^{n}-1)}{9}-n \right )$
$\displaystyle =\frac{50}{81}(10^{n}-1)-\frac{5n}{9}$
(ii) .6+.66+.666+...
Let $\displaystyle S_{n}=.6+.66+.666+...to \: n\: terms$
Let $\displaystyle =6(0.1+0.11+0.111+...to \: n\: terms$
$\displaystyle =\frac{6}{9}\left (0.9+0.99+0.999+...to \: n\: terms \right )$
$\displaystyle =\frac{6}{9}\left [ \left ( 1-\frac{1}{10} \right ) + \left ( 1-\frac{1}{10^{2}} \right )+ \left ( 1-\frac{1}{10^{3}} \right )+to \: n\: terms \right]$
$\displaystyle =\frac{2}{3}\left [ (1+1+to \: n\: terms )-\left ( \frac{1}{10}+\frac{1}{10^{2}}+\frac{1}{10^{3}}+to \: n\: terms \right ) \right]$
$\displaystyle =\frac{2}{3}\left [ n-\frac{1}{10}\left ( 1+\frac{1}{10}+\frac{1}{10^{2}}+to \: n\: terms \right ) \right]$
$\displaystyle =\frac{2}{3}\left [ n-\frac{1}{10}\left ( \frac{1-\frac{1}{10^{n}}}{1-\frac{1}{10}} \right ) \right]$
$\displaystyle =\frac{2}{3}\left [ n-\frac{1}{10}\left ( \frac{1-\frac{1}{10^{n}}}{9/10} \right ) \right]$
$\displaystyle =\frac{2}{3}\left [ n-\left ( \frac{1-10^{-n}}{9} \right ) \right]$
$\displaystyle =\frac{2}{3}n-\frac{2}{27}(1-10^{-n})$
22. Find the $\displaystyle 20^{th}$ term of the series $\displaystyle 2\times 4+4\times 6+6\times 8+..+ n\; terms$
Given the series $\displaystyle 2\times 4+4\times 6+6\times 8+..+ n\; terms$
therefore $\displaystyle n^{th}$ term is $\displaystyle a_{n}=2n\times (2n+2)==4n^{2}+4n$
$\displaystyle a_{20}=4(20)^{2}+4(20)$
$\displaystyle a_{20}=4\times 400+80$
$\displaystyle a_{20}=1600+80=1680$
23. Find the sum of the first n terms of the series: 3+7+13+21+31+...
let $\displaystyle S=3+7+13+21+31+...a_{n-1}+a_{n}$
$\displaystyle S=3+7+13+21+...a_{n-2}+a_{n-1}+a_{n}$
On subtracting both equations,
$\displaystyle S-S=(3+7+13+21+...+a_{n-1}+a_{n})-(3+7+13+21+...a_{n-2}+a_{n-1}+a_{n})$
$\displaystyle 0=3+[(7-3)+(13-7)+(21-13)+...+(a_{n}-a_{n-1})]-a_{n}$
$\displaystyle 0=3+[4+6+8+...+(n-1) terms]-a_{n}$
$\displaystyle a_{n}=3+[4+6+8+...+(n-1) terms]$
$\displaystyle a_{n}=3+\frac{n-1}{2}[2\times 4+(n-1-1)\times 2]$
$\displaystyle =3+\frac{n-1}{2}[8+(n-2)\times 2]$
$\displaystyle =3+\frac{n-1}{2}[8+2n-4]$
$\displaystyle =3+\frac{n-1}{2}[4+2n]$
$\displaystyle =3+(n-1)(n+2)$ $\displaystyle =3+n^{2}+2n-n-2=n^{2}+n+1$
$\displaystyle S_{n}=\sum_{k=1}^{n}a_{k}=\sum_{k=1}^{n}k^{2}+k+1$
$\displaystyle =\sum_{k=1}^{n}k^{2}+\sum_{k=1}^{n}k+\sum_{k=1}^{n}1$
$\displaystyle =\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}+n$
$\displaystyle =n\left [\frac{(n+1)(2n+1)}{6}+\frac{(n+1)}{2}+1 \right ]$
$\displaystyle =n\left [\frac{2n^{2}+2n+1+n}{6}+\frac{(n+1)}{2}+1 \right ]$
$\displaystyle =n\left [\frac{2n^{2}+3n+1+3n+3+6}{6} \right ]$
$\displaystyle =n\left [\frac{2n^{2}+6n+10}{6} \right ]$
$\displaystyle =n\left [\frac{n^{2}+3n+5}{3} \right ]$
24. If $\displaystyle S_{1},S_{2},S_{3}$ are the sum of first n natural numbers, their squares and their cubes respectively, show that $\displaystyle 9S_{2}^{2}=S_{3}(1+8S_{1})$
Given that,
$\displaystyle S_{1}=\frac{n(n+1)}{2}$
$\displaystyle S_{2}=\frac{n(n+1)(2n+1)}{6}$
$\displaystyle S_{3}=\frac{n^{2}(n+1)^{2}}{4}$
To prove $\displaystyle 9S_{2}^{2}=S_{3}(1+8S_{1})$,
consider L.H.S.
$\displaystyle 9S_{2}=9\left [\frac{n(n+1)(2n+1)}{6} \right ]^{2}$
$\displaystyle 9S_{2}=9\left [\frac{n^{2}(n+1)^{2}(2n+1)^{2}}{36} \right ]$
$\displaystyle 9S_{2}=\frac{n^{2}(n+1)^{2}(2n+1)^{2}}{4}$ ...(i)
consider R.H.S.
$\displaystyle S_{3}(1+8S_{1})=\frac{n^{2}(n+1)^{2}}{4}\left [ 1+8\left ( \frac{n(n+1)}{2} \right ) \right ]$
$\displaystyle =\frac{n^{2}(n+1)^{2}}{4}\left [ 1+4n(n+1) \right ]$
$\displaystyle =\frac{n^{2}(n+1)^{2}}{4}\left [ 1+4n^{2}+4n \right ]$
$\displaystyle =\frac{n^{2}(n+1)^{2}}{4}\left [ (2n+1)^{2} \right ]$
$\displaystyle S_{3}(1+8S_{1})=\frac{n^{2}(n+1)^{2}(2n+1)^{2} }{4}$ ...(ii)
from (i) and (ii), L.H.S.=R.H.S.
25. Find the sum of the following series up to n terms $\displaystyle \frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+...$
From the series $\displaystyle n^{th}$ term is $\displaystyle \frac{1^{3}+2^{3}+3^{3}+...+n^{3}}{1+3+5+...+(2n-1)}$
$\displaystyle =\frac{\left ( \frac{n^{2}(n+1)^{2}}{4} \right )}{n^{2}}$
$\displaystyle = \frac{n^{2}(n+1)^{2}}{4n^{2}}=\frac{(n+1)^{2}}{4}$
$\displaystyle =\frac{n^{2}+1+2n}{4}$
$\displaystyle =\frac{1}{4}n^{2}+\frac{1}{2}n+\frac{1}{4}$
$\displaystyle S_{n}=\sum_{k=1}^{n}a_{k}=\sum_{k=1}^{n}(\frac{1}{4}k^{2}+\frac{1}{2}k+\frac{1}{4})$
$\displaystyle =\frac{1}{4}\sum_{k=1}^{n}k^{2}+\frac{1}{2}\sum_{k=1}^{n}k+\frac{1}{4}\sum_{k=1}^{n}1$
$\displaystyle =\frac{1}{4}\left [ \frac{n(n+1)(2n+1)}{6} \right ]+\frac{1}{2}\left [ \frac{n(n+1)}{2} \right ]+\frac{1}{4}n$
$\displaystyle =\frac{n(n+1)(2n+1)}{24} + \frac{n(n+1)}{4} +\frac{n}{4}$
$\displaystyle =\frac{n[(n+1)(2n+1)+6(n+1)+6]}{24}$
$\displaystyle =\frac{n[2n^{2}+n+2n+1+6n+6+6]}{24}$
$\displaystyle =\frac{n[2n^{2}+9n+13]}{24}$
26. Show that $\displaystyle \frac{1\times 2^{2}+2\times 3^{2}+...n\times (n+1)^{2}}{1^{2}\times 2+2^{2}\times 3+...+n^{2}\times (n+1)}=\frac{3n+5}{3n+1}$
consider numerator, $\displaystyle 1\times 2^{2}+2\times 3^{2}+...n\times (n+1)^{2}$
$\displaystyle n^{th}$ term is $\displaystyle n(n+1)^{2}=n(n^{2}+1+2n)=n^{3}+2n^{2}+n$
$\displaystyle S_{n}=\sum_{k=1}^{n}a_{k}=\sum_{k=1}^{n}(k^{3}+2k^{2}+k)$
$\displaystyle =\sum_{k=1}^{n}k^{3}+2\sum_{k=1}^{n}k^{2}+\sum_{k=1}^{n}k$
$\displaystyle =\frac{n^{2}(n+1)^{2}}{4}+2\left ( \frac{n(n+1)(2n+1)}{6} \right )+\frac{n(n+1)}{2}$
$\displaystyle =\frac{n^{2}(n+1)^{2}}{4}+ \frac{2n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}$
$\displaystyle =\frac{n(n+1)}{2}\left [\frac{n(n+1)}{2}+ \frac{2(2n+1)}{3}+1 \right ]$
$\displaystyle =\frac{n(n+1)}{2}\left [\frac{n^{2}+n}{2}+ \frac{4n+2}{3}+1 \right ]$
$\displaystyle =\frac{n(n+1)}{2}\left [\frac{3n^{2}+3n+8n+4+6}{6} \right ]$
$\displaystyle =\frac{n(n+1)}{2}\left [\frac{3n^{2}+11n+10}{6} \right ]$
$\displaystyle =\frac{n(n+1)}{12}[(n+2)(3n+5)]$
$\displaystyle =\frac{n(n+1)(n+2)(3n+5)}{12}$ ...(i)
consider denominator, $\displaystyle 1^{2}\times 2+2^{2}\times 3+...+n^{2}\times (n+1)$
$\displaystyle n^{th}$ term is $\displaystyle n^{2}(n+1)=n^{3}+n^{2}$
$\displaystyle S_{n}=\sum_{k=1}^{n}a_{k}=\sum_{k=1}^{n}(k^{3}+k^{2})$
$\displaystyle =\sum_{k=1}^{n}k^{3}+\sum_{k=1}^{n}k^{2}$
$\displaystyle =\frac{n^{2}(n+1)^{2}}{4}+\frac{n(n+1)(2n+1)}{6}$
$\displaystyle =\frac{n(n+1)}{2}\left [\frac{n(n+1)}{2}+ \frac{2n+}{3} \right ]$
$\displaystyle =\frac{n(n+1)}{2}\left [\frac{n^{2}+n}{2}+ \frac{2n+}{3} \right ]$
$\displaystyle =\frac{n(n+1)}{2}\left [\frac{3n^{2}+3n+4n+2}{6} \right ]$
$\displaystyle =\frac{n(n+1)}{2}\left [\frac{3n^{2}+7n+2}{6} \right ]$
$\displaystyle =\frac{n(n+1)}{12}(3n^{2}+7n+2)$
$\displaystyle =\frac{n(n+1)}{12}(n+2)(3n+1)$
$\displaystyle =\frac{n(n+1)(n+2)(3n+1)}{12}$ ...(ii)
Using (i) and (ii),
$\displaystyle \frac{1\times 2^{2}+2\times 3^{2}+...n\times (n+1)^{2}}{1^{2}\times 2+2^{2}\times 3+...+n^{2}\times (n+1)}$
$\displaystyle =\frac{\frac{n(n+1)(n+2)(3n+5)}{12}}{\frac{n(n+1)(n+2)(3n+1)}{12}}$
$\displaystyle =\frac{(3n+5)}{(3n+1)}$
27. A farmer buys a used tractor for Rs. 12000. He pays Rs. 6000 cash and agrees to pay the balance in annual instalments of Rs. 500 plus 12% interest on the unpaid amount. How much will the tractor cost him?
Given, cost of tractor = 12000
amount paid = 6000
amount unpaid = 12000-6000 = 6000
Interest paid annually is,
12% of 6000, 12% of 5500, 12% of 5000,...12% of 500
Total interest to be paid is
=12% of 6000+12% of 5500+12% of 5000+...+12% of 500
=12%[6000+5500+5000+...+500]
=12%[500+1000+...+6000] ...(i)
consider the series 500,1000,...,6000
let the number of terms be n and a=500, l=6000, d=500
We know that,
l=a+(n-1)d
6000=500+(n-1)500
6000=500[1+n-1]
n=12
$\displaystyle S_{n}=\frac{n}{2}(2a+(n-1)d)$
$\displaystyle =\frac{12}{2}(2(500)+(12-1)500)$
$\displaystyle =6(1000+11 \times 500)$
$\displaystyle =6(1000+5500)=39000$
Interest to be paid is 12%[500+1000+...+6000]=12% (39000)=4680
Total cost of tractor = 12000+4680=16680
28. Shamshad Ali buys a scooter for Rs. 22000. He pays Rs. 4000 cash and agrees to pay the balance in annual instalment of Rs. 1000 plus 10% interest on the unpaid amount. How much will the tractor cost him?
Given, cost of scooter = 22000
amount paid = 4000
amount unpaid = 22000-4000 = 18000
Interest paid annually is,
10% of 18000, 10% of 17000, 10% of 16000,...10% of 1000
Total interest to be paid is
=10% of 18000+10% of 17000+10% of 16000,...+10% of 1000
=10%[18000+17000+...+1000]
=10%[1000+2000+...+18000] ...(i)
consider the series 1000,2000,...,18000
let the number of terms be n and a=1000, l=18000, d=1000
We know that,
l=a+(n-1)d
18000=1000+(n-1)1000
18000=1000[1+n-1]
n=18
$\displaystyle S_{n}=\frac{n}{2}(2a+(n-1)d)$
$\displaystyle =\frac{18}{2}(2(1000)+(18-1)1000)$
$\displaystyle =9(2000+17 \times 1000)$
$\displaystyle =9(2000+17000)=171000$
Interest to be paid is 10%[1000+2000+...+18000]=10% (171000)=17100
Total cost of tractor = 22000+17100=39100
29. A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8th set of letter is mailed.
Number of letters mailed is $\displaystyle 4,4^{2},4^{3},...4^{8}$
a=4, r=4, n=8
$\displaystyle S_{n}=\frac{a(r^{n}-1)}{r-1}$
$\displaystyle =\frac{4(4^{8}-1)}{4-1}$
$\displaystyle =\frac{4(65536-1)}{3}$
$\displaystyle =\frac{4(65535)}{3}=87380$
Given the cost of one mail is 50paise,
Cost of 87380 is $\displaystyle =87380\frac{50}{100}=43690$
30. A man deposited Rs. 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years.
Amount deposited = 10000
rate of interest = 5%
interest amount = 5% of 10000 = 500 Rs
Amount in 15th year is $\displaystyle =10000+14 \times 500 = 17000$
Amount after 20 year is $\displaystyle =10000+20 \times 500 = 20000$
31. A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.
Cost of machine = 15625 Rs
Machine value depreciates by 20% every year.
therefore, its value after every year is 80% less than originial cost
⇒ $\displaystyle \frac{80}{100}=\frac{4}{5}$
Value at the end of 5 years is
$\displaystyle =5\times \frac{4}{5}\times 15625=5120$
32. 150 workers were engaged to finish a job in a certain number of days, 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.
Let x be the number of days in which 150 workers finish the work.
Given that,
150x=150+146+142+.. (x+8) terms
the series 150+146+142... is an A.P.
a=150, d=-4, n=x+8
$\displaystyle 150x=\frac{x+8}{2}[2(150)+(x+8-1)(-4)]$
$\displaystyle 150x=(x+8)[150+(x+7)(-2)]$
$\displaystyle 75x=(x+8)[75+(x+7)(-1)]$
$\displaystyle 75x=(x+8)[75-x-7]$
$\displaystyle 75x=(x+8)[68-x]$
$\displaystyle 75x=68x-x^{2}+544-8x$
$\displaystyle x^{2}-135x-544=0$
$\displaystyle (x-17)(x+32)=0$
$\displaystyle x=17$ or $\displaystyle x=-32$
x cannot be negative therefore $\displaystyle x=17$
The number of days required= 17+8=25
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