1. Find the sum of odd integers from 1 to 2001
Odd integers from 1 to 2001 are 1,3,5....2001
$\displaystyle a=1$
$\displaystyle d=3-1=2$
$\displaystyle a_{n}=l=2001$
We know that,
$\displaystyle a_{n}=a+(n-1)d$
$\displaystyle 2001=1+(n-1)2$
$\displaystyle 2001=2n-1$
$\displaystyle 2002=2n$
$\displaystyle n=1001$
sum is given by $\displaystyle S_{n}=\frac{n}{2}(a+l)$
$\displaystyle S_{1001}=\frac{1001}{2}(1+2001)$
$\displaystyle S_{1001}=1001*1001=1002001$
2. Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.
Natural numbers between 100 and 1000, which are multiples of 5 are 105,110...995
$\displaystyle a=105$
$\displaystyle d=110-105=5$
$\displaystyle a_{n}=l=995$
We know that,
$\displaystyle a_{n}=a+(n-1)d$
$\displaystyle 995=105+(n-1)5$
$\displaystyle 890=(n-1)5$
$\displaystyle 178=n-1$
$\displaystyle n=179$
sum is given by $\displaystyle S_{n}=\frac{n}{2}(a+l)$
$\displaystyle S_{179}=\frac{179}{2}(105+995)$
$\displaystyle S_{179}=179*550=98450$
3. In an A.P. , the first term is 2 and sum of the first five terms is one-fourth of the next five terms. Show that 20th term is -112.
Given, a=2
sum of first five terms=one-fourth of sum of next five terms
$\displaystyle S_{5}=\frac{1}{4}(S_{10}-S_{5})$
$\displaystyle 4S_{5}=S_{10}-S_{5}$
$\displaystyle 5S_{5}=S_{10}$
$\displaystyle 5[\frac{5}{2}(2*2+(5-1)d)]$ $\displaystyle =\frac{10}{2}(2*2+(10-1)d)$ [$\displaystyle S_{n}=\frac{n}{2}(2a+(n-1)d)$]
$\displaystyle \frac{25}{2}(4+4d)=\frac{10}{2}(4+9d)$
$\displaystyle 25(4+4d)=10(4+9d)$
$\displaystyle 5(4+4d)=2(4+9d)$
$\displaystyle 20+20d=8+18d$
$\displaystyle 2d=-12$
$\displaystyle d=-6$
We know that $\displaystyle a_{n}=a+(n-1)d$
Put n=20 we get
$\displaystyle a_{20}=2+(20-1)(-6)$
$\displaystyle a_{20}=2-114=-112$
4. How many terms of A.P. $\displaystyle -6,\frac{-11}{2},-5,...$ are needed to give the sum -25?
Given $\displaystyle a=-6, S_{n}=-25$
$\displaystyle d=\frac{-11}{2}-(-6)=\frac{1}{2}$
We know that, $\displaystyle S_{n}=\frac{n}{2}(2a+(n-1)d)$
$\displaystyle -25=\frac{n}{2}(2(-6)+(n-1)\frac{1}{2})$
$\displaystyle -50=n(-12+\frac{n-1}{2})$
$\displaystyle -50=n(\frac{-24+n-1}{2})$
$\displaystyle -100=n(n-25)$
$\displaystyle n^{2}-25n+100=0$
$\displaystyle (n-20)(n-5)=0$
$\displaystyle n=20$ or $\displaystyle n=5$
5. In an A.P. if $\displaystyle p^{th}$ term is $\displaystyle \frac{1}{q}$ and $\displaystyle q^{th}$ term is $\displaystyle \frac{1}{p}$, prove that the sum of first $\displaystyle pq$ terms is $\displaystyle \frac{1}{2}(pq+1)$, where $\displaystyle p\neq q$
Given $\displaystyle a_{p}=\frac{1}{q}$
$\displaystyle a_{q}=\frac{1}{p}$
we know that, $\displaystyle a_{n}=a+(n-1)d$
$\displaystyle a_{p}=a+(p-1)d=\frac{1}{q}$... (i)
$\displaystyle a_{q}=a+(q-1)d=\frac{1}{p}$... (ii)
Subtract (ii) from (i)
$\displaystyle a+(p-1)d-(a+(q-1)d)$ $\displaystyle =\frac{1}{q}-\frac{1}{p}$
$\displaystyle (p-q)d=\frac{p-q}{pq}$
$\displaystyle d=\frac{1}{pq}$
Substitute value of d in (i)
$\displaystyle a+(p-1)\frac{1}{pq}=\frac{1}{q}$
$\displaystyle a+\frac{1}{q}-\frac{1}{pq}=\frac{1}{q}$
$\displaystyle a=\frac{1}{pq}$
Now, $\displaystyle S_{n}=\frac{n}{2}(2a+(n-1)d)$
$\displaystyle S_{pq}=\frac{pq}{2}(2\frac{1}{pq}+(pq-1)\frac{1}{pq})$
$\displaystyle =\frac{pq}{2}(\frac{2+pq-1}{pq})$
$\displaystyle =\frac{1}{2}(1+pq)$
6. If the sum of a certain number of terms of the A.P. 25,22,19,... is 116. Find the last term.
Given $\displaystyle a=25$
$\displaystyle d=22-25=-3$
$\displaystyle S_{n}=116$
We know that, $\displaystyle S_{n}=\frac{n}{2}(2a+(n-1)d)$
$\displaystyle 116=\frac{n}{2}(2*25+(n-1)(-3))$
$\displaystyle 116=\frac{n}{2}(50-3n+3)$
$\displaystyle 232=n(53-3n)$
$\displaystyle 3n^{2}-53n+232=0$
$\displaystyle n=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$
$\displaystyle n=\frac{-53\pm \sqrt{(53)^{2}-4*3*232}}{2*3}$
$\displaystyle n=\frac{-53\pm \sqrt{2809-2784}}{6}$
$\displaystyle n=\frac{-53\pm \sqrt{25}}{6}=\frac{-53\pm 5}{6}$
$\displaystyle n=\frac{58}{6}$ or $\displaystyle n=\frac{48}{6}$
$\displaystyle n=\frac{29}{3}$ or $\displaystyle n=8$
$\displaystyle n=\frac{29}{3}$ is not possible
∴ $\displaystyle n=8$
the last term is $\displaystyle a_{n}=a+(n-1)d$
$\displaystyle a_{8}=25+(8-1)(-3)$
$\displaystyle a_{8}=25-21=4$
7. Find the sum to n terms of A.P. , whose $\displaystyle k^{th}$ term is 5k+1.
Given $\displaystyle a_{k}=5k+1$
Put k=1 and k=n
$\displaystyle a_{1}=6$
$\displaystyle a_{n}=l=5n+1$
We know that, $\displaystyle S_{n}=\frac{n}{2}(a+l)$
$\displaystyle S_{n}=\frac{n}{2}(6+5n+1)=\frac{n}{2}(5n+7)$
8. If the sum of n terms of an A.P. is $\displaystyle (pn+qn^{2})$, where p and q are constants, find the common difference.
Given, $\displaystyle S_{n}=pn+qn^{2}$
We know that, $\displaystyle a_{n}=S_{n}-S_{n-1}$
$\displaystyle =pn+qn^{2}-(p(n-1)+q(n-1)^{2})$
$\displaystyle =pn+qn^{2}-pn+p$ $\displaystyle -q(n^{2}+1-2n)$
$\displaystyle =qn^{2}+p-qn^{2}-q+2qn$
$\displaystyle a_{n}=p-q+2qn$
Now, $\displaystyle d=a_{n}-a_{n-1}$
$\displaystyle =p-q+2qn-(p-q+2q(n-1))$
$\displaystyle =p-q+2qn-p+q-2qn+2q$
$\displaystyle d=2q$
9. The sum of n terms of two arithmetic progressions are in the ratio $\displaystyle 5n+4:9n+6$. Find the ratio of their $\displaystyle 18^{th}$ terms.
Let $\displaystyle a_{1}, a_{2}$ and $\displaystyle d_{1}, d_{2}$ be the first terms and common differences of two A.P.s respectively.
Given, $\displaystyle \frac{sum of 1^{st} A.P.}{sum of 2^{nd} A.P.}=\frac{5n+4}{9n+6}$
$\displaystyle \frac{\frac{n}{2}(2a_{1}+(n-1)d_{1})}{\frac{n}{2}(2a_{2}+(n-1)d_{2})}=\frac{5n+4}{9n+6}$
$\displaystyle \frac{2a_{1}+(n-1)d_{1}}{2a_{2}+(n-1)d_{2}}=\frac{5n+4}{9n+6}$ ...(i)
Consider,
$\displaystyle =\frac{a_{1}+17d_{1}}{a_{2}+17d_{2}}$ $\displaystyle =\frac{2a_{1}+34d_{1}}{2a_{2}+34d_{2}}$ ...(ii)
Comparing (i) and (ii),
$\displaystyle n-1=34$ ⇒ $\displaystyle n=35$
(i) ⇒ $\displaystyle \frac{2a_{1}+34d_{1}}{2a_{2}+34d_{2}}=\frac{5(35)+4}{9(35)+6}$
$\displaystyle \frac{18^{th} term of 1^{st} A.P.}{18^{th} term of 2^{nd} A.P.}=\frac{175+4}{315+6}$ $\displaystyle =\frac{179}{321}$
10. If the sum of first p terms of an A.P. is equal to the sum of first q terms, then find the sum of the first (p+q) terms.
Given, $\displaystyle S_{p}=S_{q}$
$\displaystyle \frac{p}{2}(2a+(p-1)d)$ $\displaystyle =\frac{q}{2}(2a+(q-1)d)$
$\displaystyle 2ap+p^{2}d-pd=2aq+q^{2}d-qd$
$\displaystyle 2ap-2aq=q^{2}d-qd-p^{2}d+pd$
$\displaystyle 2a(p-q)=[-(p^{2}-q^{2})+(p-q)]d$
$\displaystyle 2a(p-q)=[-(p-q)(p+q)$ $\displaystyle +(p-q)]d$
$\displaystyle 2a(p-q)=(p-q)[-(p+q)+1]d$
$\displaystyle 2a=(1-p-q)d$ ...(i)
Now, $\displaystyle S_{p+q}=\frac{p+q}{2}(2a+(p+q-1)d)$
$\displaystyle S_{p+q}=\frac{p+q}{2}(2a-(1-p-q)d)$
$\displaystyle S_{p+q}=\frac{p+q}{2}(2a-2a)=0$
11. Sum of the p, q and r terms of an A.P. are a,b and c respectively. Prove that $\displaystyle \frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0$
Let A and d be first term and common difference of an A.P.
Given,
$\displaystyle S_{p}=a$
$\displaystyle \frac{p}{2}(2A+(p-1)d)=a$
$\displaystyle A+\frac{(p-1)d}{2}=\frac{a}{p}$ ...(i)
$\displaystyle S_{q}=b$
$\displaystyle \frac{q}{2}(2A+(q-1)d)=b$
$\displaystyle A+\frac{(q-1)d}{2}=\frac{b}{q}$ ...(ii)
$\displaystyle S_{r}=c$
$\displaystyle \frac{r}{2}(2A+(r-1)d)=c$
$\displaystyle A+\frac{(r-1)d}{2}=\frac{c}{r}$ ...(iii)
Consider,
$\displaystyle \frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0$
Substituting (i),(ii),(iii) ⇒
$\displaystyle [A+\frac{(p-1)d}{2}](q-r)$ $\displaystyle +[A+\frac{(q-1)d}{2}](r-p)$ $\displaystyle +[A+\frac{(r-1)d}{2}](p-q)]=0$
$\displaystyle Aq+\frac{(p-1)qd}{2}-Ar-\frac{(p-1)rd}{2}$ $\displaystyle +Ar+\frac{(q-1)rd}{2}-Ap-\frac{(q-1)pd}{2}$ $\displaystyle +Ap+\frac{(r-1)pd}{2}-Aq-\frac{(r-1)qd}{2}=0$
$\displaystyle \frac{(p-1)qd}{2}-\frac{(p-1)rd}{2}$ $\displaystyle +\frac{(q-1)rd}{2}-\frac{(q-1)pd}{2}$ $\displaystyle +\frac{(r-1)pd}{2}-\frac{(r-1)qd}{2}=0$
$\displaystyle \frac{d}{2}[pq-q-pr+r+qr$ $\displaystyle -r-qp+p+pr-p-qr+q]=0$
$\displaystyle \frac{d}{2}[0]=0$
$\displaystyle 0=0$
L.H.S.=R.H.S.
12. The ratio of the sums of m and n terms of an A.P. is $\displaystyle m^{2}:n^{2}$. Show the ration of $\displaystyle m^{th}$ and $\displaystyle n^{th}$ term is $\displaystyle (2m-1):(2n-1)$
Given,
$\displaystyle \frac{S_{m}}{S_{n}}=\frac{m^{2}}{n^{2}}$
$\displaystyle \large \frac{\frac{m}{2}(2a+(m-1)d)}{\frac{n}{2}(2a+(n-1)d)}=\frac{m^{2}}{n^{2}}$
$\displaystyle \frac{m(2a+(m-1)d)}{n(2a+(n-1)d)}=\frac{m^{2}}{n^{2}}$
$\displaystyle \frac{2a+(m-1)d}{2a+(n-1)d}=\frac{m}{n}$
$\displaystyle n(2a+(m-1)d)$ $\displaystyle =m(2a+(n-1)d$
$\displaystyle 2an+mnd-nd$ $\displaystyle =2am+mnd-md$
$\displaystyle 2an-nd=2am-md$
$\displaystyle 2an-2am=-md+nd$
$\displaystyle 2a(n-m)=d(n-m)$
$\displaystyle d=2a$
Now, $\displaystyle \frac{a_{m}}{a_{n}}=\frac{a+(m-1)d}{a+(n-1)d}$
$\displaystyle \frac{a_{m}}{a_{n}}=\frac{a+(m-1)2a}{a+(n-1)2a}$
$\displaystyle =\frac{a+(m-1)2a}{a+(n-1)2a}$
$\displaystyle =\frac{a+2am-2a}{a+2an-2a}$
$\displaystyle =\frac{2am-a}{2an-a}=\frac{a(2m-1)}{a(2n-1)}$
$\displaystyle \frac{a_{m}}{a_{n}}=\frac{2m-1}{2n-1}$
13. If the sum of n terms of an A.P. is $\displaystyle 3n^{2}+5n$ and its $\displaystyle m^{th}$ term is 164, find the value of m.
Given, $\displaystyle S-{n}=3n^{2}+5n$
$\displaystyle a_{m}=164$
We know that,
$\displaystyle a_{n}=S_{n}-S_{n-1}$
$\displaystyle a_{n}=3n^{2}+5n-(3(n-1)^{2}$ $\displaystyle +5(n-1))$
$\displaystyle =3n^{2}+5n-(3(n^{2}+1-2n)$ $\displaystyle +5n-5)$
$\displaystyle =3n^{2}+5n-3n^{2}-3$ $\displaystyle +6n-5n+5$
$\displaystyle a_{n}=6n+2$
Now, $\displaystyle a_{m}=6m+2=164$
$\displaystyle 6m=162$
$\displaystyle m=27$
14. Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.
Let $\displaystyle A_{1}, A_{2},A_{3}, A_{4},A_{5}$ are the five numbers between 8 and 26 such that $\displaystyle 8,A_{1}, A_{2},A_{3}, A_{4},A_{5},26$ are in A.P.
$\displaystyle a=8, l=26, n=7$
We know that, $\displaystyle a_{n}=l=a+(n-1)d$
$\displaystyle 26=8+(7-1)d$
$\displaystyle 18=6d$
$\displaystyle d=3$
Now,
$\displaystyle A_{1}=a+d=8+3=11$
$\displaystyle A_{2}=a+2d=8+6=14$
$\displaystyle A_{3}=a+3d=8+9=17$
$\displaystyle A_{4}=a+4d=8+12=20$
$\displaystyle A_{5}=a+5d=8+15=23$
Hence, five numbers between 8 and 26 are 11,14,17,20,23
15. If $\displaystyle \frac{a^{n}+b^{n}}{a^{n-1}+b^{n-1}}$ is the A.M. between a and b, then find the value of n.
Given $\displaystyle A=\frac{a^{n}+b^{n}}{a^{n-1}+b^{n-1}}$
We know that, $\displaystyle A=\frac{a+b}{2}$
$\displaystyle \frac{a+b}{2}=\frac{a^{n}+b^{n}}{a^{n-1}+b^{n-1}}$
$\displaystyle (a+b)(a^{n-1}+b^{n-1})=2(a^{n}+b^{n})$
$\displaystyle a^{n}+ab^{n-1}+ba^{n-1}+b^{n}$ $\displaystyle =2(a^{n}+b^{n})$
$\displaystyle ab^{n-1}+ba^{n-1}=a^{n}+b^{n}$
$\displaystyle ab^{n-1}-b^{n}=a^{n}-+ba^{n-1}$
$\displaystyle b^{n-1}(a-b)=a^{n-1}(a-b)$
$\displaystyle b^{n-1}=a^{n-1}$
$\displaystyle \frac{a^{n-1}}{b^{n-1}}=1$
$\displaystyle (\frac{a}{b})^{n-1}=(\frac{a}{b})^{0}$
$\displaystyle n-1=0$
$\displaystyle n=1$
16. Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of $\displaystyle 7^{th}$ and $\displaystyle (m-1)^{th}$ numbers is $\displaystyle 5:9$. Find the value of m.
Let $\displaystyle A_{1}, A_{2},A_{3},...A_{m}$ are m numbers between 1 and 31.
Given, $\displaystyle a=1, l=a_{m+2}=31$
We know that,
$\displaystyle l=a+(n-1)d$
$\displaystyle 31=1+(m+2-1)d$
$\displaystyle 30=(m+1)d$
$\displaystyle d=\frac{30}{m+1}$
Given, $\displaystyle \frac{A_{7}}{A_{m-1}}=\frac{5}{9}$
$\displaystyle \frac{a+7d}{a+(m-1)d}=\frac{5}{9}$
$\displaystyle \large \frac{1+7\frac{30}{m+1}}{1+(m-1)\frac{30}{m+1}}=\frac{5}{9}$
$\displaystyle \frac{m+1+210}{m+1+30m-30}=\frac{5}{9}$
$\displaystyle \frac{m+1+210}{31m-29}=\frac{5}{9}$
$\displaystyle 9m+1899=155m-145$
$\displaystyle 146m=2044$
$\displaystyle m=14$
17. A man starts repaying a loan as first installment of Rs.100. If he increases the installment by Rs.5 every month, what amount he will pay in the $\displaystyle 30^{th}$ installment?
Given, first installment=100, second installment=105 and so on...
$\displaystyle a=100, d=105-100=5$
We know that, $\displaystyle a_{n}=a+(n-1)d$
$\displaystyle 30^{th}$ installment is
$\displaystyle a_{30}=100+(30-1)5$
$\displaystyle a_{30}=100+29*5$
$\displaystyle a_{30}=100+145=245$
18. The difference between any two consecutive interior angles of a polygon is $\displaystyle 5^{0}$. If the smallest angle is $\displaystyle 120^{0}$, find the number of the sides of the polygon.
Let the number of sides of the polygon be n.
Given, $\displaystyle a=120, d=5$
We know that, sum of all interior angles of a polygon is $\displaystyle (n-2)180^{0}$
∴ $\displaystyle S_{n}=(n-2)180^{0}$
$\displaystyle \frac{n}{2}(2a+(n-1)d)=(n-2)180$
$\displaystyle \frac{n}{2}(2*120+(n-1)5)=(n-2)180$
$\displaystyle n(240+5n-5)=2(n-2)180$
$\displaystyle n(235+5n)=(n-2)360$
$\displaystyle 235n+5n^{2}=360n-720$
$\displaystyle 5n^{2}-125n+720=0$
$\displaystyle n^{2}-25n+144=0$
$\displaystyle (n-16)(n-9)=0$
$\displaystyle n=16$ or $\displaystyle n=9$
We know that, $\displaystyle a_{n}=a+(n-1)d$
$\displaystyle n=16$ ⇒ $\displaystyle a_{16}=120+15*5=195^{0}>180^{0}$ which is not possible.
$\displaystyle n=9$ ⇒ $\displaystyle a_{9}=120+8*5=160^{0}<180^{0}$
∴ $\displaystyle n=9$
Odd integers from 1 to 2001 are 1,3,5....2001
$\displaystyle a=1$
$\displaystyle d=3-1=2$
$\displaystyle a_{n}=l=2001$
We know that,
$\displaystyle a_{n}=a+(n-1)d$
$\displaystyle 2001=1+(n-1)2$
$\displaystyle 2001=2n-1$
$\displaystyle 2002=2n$
$\displaystyle n=1001$
sum is given by $\displaystyle S_{n}=\frac{n}{2}(a+l)$
$\displaystyle S_{1001}=\frac{1001}{2}(1+2001)$
$\displaystyle S_{1001}=1001*1001=1002001$
2. Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.
Natural numbers between 100 and 1000, which are multiples of 5 are 105,110...995
$\displaystyle a=105$
$\displaystyle d=110-105=5$
$\displaystyle a_{n}=l=995$
We know that,
$\displaystyle a_{n}=a+(n-1)d$
$\displaystyle 995=105+(n-1)5$
$\displaystyle 890=(n-1)5$
$\displaystyle 178=n-1$
$\displaystyle n=179$
sum is given by $\displaystyle S_{n}=\frac{n}{2}(a+l)$
$\displaystyle S_{179}=\frac{179}{2}(105+995)$
$\displaystyle S_{179}=179*550=98450$
3. In an A.P. , the first term is 2 and sum of the first five terms is one-fourth of the next five terms. Show that 20th term is -112.
Given, a=2
sum of first five terms=one-fourth of sum of next five terms
$\displaystyle S_{5}=\frac{1}{4}(S_{10}-S_{5})$
$\displaystyle 4S_{5}=S_{10}-S_{5}$
$\displaystyle 5S_{5}=S_{10}$
$\displaystyle 5[\frac{5}{2}(2*2+(5-1)d)]$ $\displaystyle =\frac{10}{2}(2*2+(10-1)d)$ [$\displaystyle S_{n}=\frac{n}{2}(2a+(n-1)d)$]
$\displaystyle \frac{25}{2}(4+4d)=\frac{10}{2}(4+9d)$
$\displaystyle 25(4+4d)=10(4+9d)$
$\displaystyle 5(4+4d)=2(4+9d)$
$\displaystyle 20+20d=8+18d$
$\displaystyle 2d=-12$
$\displaystyle d=-6$
We know that $\displaystyle a_{n}=a+(n-1)d$
Put n=20 we get
$\displaystyle a_{20}=2+(20-1)(-6)$
$\displaystyle a_{20}=2-114=-112$
4. How many terms of A.P. $\displaystyle -6,\frac{-11}{2},-5,...$ are needed to give the sum -25?
Given $\displaystyle a=-6, S_{n}=-25$
$\displaystyle d=\frac{-11}{2}-(-6)=\frac{1}{2}$
We know that, $\displaystyle S_{n}=\frac{n}{2}(2a+(n-1)d)$
$\displaystyle -25=\frac{n}{2}(2(-6)+(n-1)\frac{1}{2})$
$\displaystyle -50=n(-12+\frac{n-1}{2})$
$\displaystyle -50=n(\frac{-24+n-1}{2})$
$\displaystyle -100=n(n-25)$
$\displaystyle n^{2}-25n+100=0$
$\displaystyle (n-20)(n-5)=0$
$\displaystyle n=20$ or $\displaystyle n=5$
5. In an A.P. if $\displaystyle p^{th}$ term is $\displaystyle \frac{1}{q}$ and $\displaystyle q^{th}$ term is $\displaystyle \frac{1}{p}$, prove that the sum of first $\displaystyle pq$ terms is $\displaystyle \frac{1}{2}(pq+1)$, where $\displaystyle p\neq q$
Given $\displaystyle a_{p}=\frac{1}{q}$
$\displaystyle a_{q}=\frac{1}{p}$
we know that, $\displaystyle a_{n}=a+(n-1)d$
$\displaystyle a_{p}=a+(p-1)d=\frac{1}{q}$... (i)
$\displaystyle a_{q}=a+(q-1)d=\frac{1}{p}$... (ii)
Subtract (ii) from (i)
$\displaystyle a+(p-1)d-(a+(q-1)d)$ $\displaystyle =\frac{1}{q}-\frac{1}{p}$
$\displaystyle (p-q)d=\frac{p-q}{pq}$
$\displaystyle d=\frac{1}{pq}$
Substitute value of d in (i)
$\displaystyle a+(p-1)\frac{1}{pq}=\frac{1}{q}$
$\displaystyle a+\frac{1}{q}-\frac{1}{pq}=\frac{1}{q}$
$\displaystyle a=\frac{1}{pq}$
Now, $\displaystyle S_{n}=\frac{n}{2}(2a+(n-1)d)$
$\displaystyle S_{pq}=\frac{pq}{2}(2\frac{1}{pq}+(pq-1)\frac{1}{pq})$
$\displaystyle =\frac{pq}{2}(\frac{2+pq-1}{pq})$
$\displaystyle =\frac{1}{2}(1+pq)$
6. If the sum of a certain number of terms of the A.P. 25,22,19,... is 116. Find the last term.
Given $\displaystyle a=25$
$\displaystyle d=22-25=-3$
$\displaystyle S_{n}=116$
We know that, $\displaystyle S_{n}=\frac{n}{2}(2a+(n-1)d)$
$\displaystyle 116=\frac{n}{2}(2*25+(n-1)(-3))$
$\displaystyle 116=\frac{n}{2}(50-3n+3)$
$\displaystyle 232=n(53-3n)$
$\displaystyle 3n^{2}-53n+232=0$
$\displaystyle n=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$
$\displaystyle n=\frac{-53\pm \sqrt{(53)^{2}-4*3*232}}{2*3}$
$\displaystyle n=\frac{-53\pm \sqrt{2809-2784}}{6}$
$\displaystyle n=\frac{-53\pm \sqrt{25}}{6}=\frac{-53\pm 5}{6}$
$\displaystyle n=\frac{58}{6}$ or $\displaystyle n=\frac{48}{6}$
$\displaystyle n=\frac{29}{3}$ or $\displaystyle n=8$
$\displaystyle n=\frac{29}{3}$ is not possible
∴ $\displaystyle n=8$
the last term is $\displaystyle a_{n}=a+(n-1)d$
$\displaystyle a_{8}=25+(8-1)(-3)$
$\displaystyle a_{8}=25-21=4$
7. Find the sum to n terms of A.P. , whose $\displaystyle k^{th}$ term is 5k+1.
Given $\displaystyle a_{k}=5k+1$
Put k=1 and k=n
$\displaystyle a_{1}=6$
$\displaystyle a_{n}=l=5n+1$
We know that, $\displaystyle S_{n}=\frac{n}{2}(a+l)$
$\displaystyle S_{n}=\frac{n}{2}(6+5n+1)=\frac{n}{2}(5n+7)$
8. If the sum of n terms of an A.P. is $\displaystyle (pn+qn^{2})$, where p and q are constants, find the common difference.
Given, $\displaystyle S_{n}=pn+qn^{2}$
We know that, $\displaystyle a_{n}=S_{n}-S_{n-1}$
$\displaystyle =pn+qn^{2}-(p(n-1)+q(n-1)^{2})$
$\displaystyle =pn+qn^{2}-pn+p$ $\displaystyle -q(n^{2}+1-2n)$
$\displaystyle =qn^{2}+p-qn^{2}-q+2qn$
$\displaystyle a_{n}=p-q+2qn$
Now, $\displaystyle d=a_{n}-a_{n-1}$
$\displaystyle =p-q+2qn-(p-q+2q(n-1))$
$\displaystyle =p-q+2qn-p+q-2qn+2q$
$\displaystyle d=2q$
9. The sum of n terms of two arithmetic progressions are in the ratio $\displaystyle 5n+4:9n+6$. Find the ratio of their $\displaystyle 18^{th}$ terms.
Let $\displaystyle a_{1}, a_{2}$ and $\displaystyle d_{1}, d_{2}$ be the first terms and common differences of two A.P.s respectively.
Given, $\displaystyle \frac{sum of 1^{st} A.P.}{sum of 2^{nd} A.P.}=\frac{5n+4}{9n+6}$
$\displaystyle \frac{\frac{n}{2}(2a_{1}+(n-1)d_{1})}{\frac{n}{2}(2a_{2}+(n-1)d_{2})}=\frac{5n+4}{9n+6}$
$\displaystyle \frac{2a_{1}+(n-1)d_{1}}{2a_{2}+(n-1)d_{2}}=\frac{5n+4}{9n+6}$ ...(i)
Consider,
$\displaystyle =\frac{a_{1}+17d_{1}}{a_{2}+17d_{2}}$ $\displaystyle =\frac{2a_{1}+34d_{1}}{2a_{2}+34d_{2}}$ ...(ii)
Comparing (i) and (ii),
$\displaystyle n-1=34$ ⇒ $\displaystyle n=35$
(i) ⇒ $\displaystyle \frac{2a_{1}+34d_{1}}{2a_{2}+34d_{2}}=\frac{5(35)+4}{9(35)+6}$
$\displaystyle \frac{18^{th} term of 1^{st} A.P.}{18^{th} term of 2^{nd} A.P.}=\frac{175+4}{315+6}$ $\displaystyle =\frac{179}{321}$
10. If the sum of first p terms of an A.P. is equal to the sum of first q terms, then find the sum of the first (p+q) terms.
Given, $\displaystyle S_{p}=S_{q}$
$\displaystyle \frac{p}{2}(2a+(p-1)d)$ $\displaystyle =\frac{q}{2}(2a+(q-1)d)$
$\displaystyle 2ap+p^{2}d-pd=2aq+q^{2}d-qd$
$\displaystyle 2ap-2aq=q^{2}d-qd-p^{2}d+pd$
$\displaystyle 2a(p-q)=[-(p^{2}-q^{2})+(p-q)]d$
$\displaystyle 2a(p-q)=[-(p-q)(p+q)$ $\displaystyle +(p-q)]d$
$\displaystyle 2a(p-q)=(p-q)[-(p+q)+1]d$
$\displaystyle 2a=(1-p-q)d$ ...(i)
Now, $\displaystyle S_{p+q}=\frac{p+q}{2}(2a+(p+q-1)d)$
$\displaystyle S_{p+q}=\frac{p+q}{2}(2a-(1-p-q)d)$
$\displaystyle S_{p+q}=\frac{p+q}{2}(2a-2a)=0$
11. Sum of the p, q and r terms of an A.P. are a,b and c respectively. Prove that $\displaystyle \frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0$
Let A and d be first term and common difference of an A.P.
Given,
$\displaystyle S_{p}=a$
$\displaystyle \frac{p}{2}(2A+(p-1)d)=a$
$\displaystyle A+\frac{(p-1)d}{2}=\frac{a}{p}$ ...(i)
$\displaystyle S_{q}=b$
$\displaystyle \frac{q}{2}(2A+(q-1)d)=b$
$\displaystyle A+\frac{(q-1)d}{2}=\frac{b}{q}$ ...(ii)
$\displaystyle S_{r}=c$
$\displaystyle \frac{r}{2}(2A+(r-1)d)=c$
$\displaystyle A+\frac{(r-1)d}{2}=\frac{c}{r}$ ...(iii)
Consider,
$\displaystyle \frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0$
Substituting (i),(ii),(iii) ⇒
$\displaystyle [A+\frac{(p-1)d}{2}](q-r)$ $\displaystyle +[A+\frac{(q-1)d}{2}](r-p)$ $\displaystyle +[A+\frac{(r-1)d}{2}](p-q)]=0$
$\displaystyle Aq+\frac{(p-1)qd}{2}-Ar-\frac{(p-1)rd}{2}$ $\displaystyle +Ar+\frac{(q-1)rd}{2}-Ap-\frac{(q-1)pd}{2}$ $\displaystyle +Ap+\frac{(r-1)pd}{2}-Aq-\frac{(r-1)qd}{2}=0$
$\displaystyle \frac{(p-1)qd}{2}-\frac{(p-1)rd}{2}$ $\displaystyle +\frac{(q-1)rd}{2}-\frac{(q-1)pd}{2}$ $\displaystyle +\frac{(r-1)pd}{2}-\frac{(r-1)qd}{2}=0$
$\displaystyle \frac{d}{2}[pq-q-pr+r+qr$ $\displaystyle -r-qp+p+pr-p-qr+q]=0$
$\displaystyle \frac{d}{2}[0]=0$
$\displaystyle 0=0$
L.H.S.=R.H.S.
12. The ratio of the sums of m and n terms of an A.P. is $\displaystyle m^{2}:n^{2}$. Show the ration of $\displaystyle m^{th}$ and $\displaystyle n^{th}$ term is $\displaystyle (2m-1):(2n-1)$
Given,
$\displaystyle \frac{S_{m}}{S_{n}}=\frac{m^{2}}{n^{2}}$
$\displaystyle \large \frac{\frac{m}{2}(2a+(m-1)d)}{\frac{n}{2}(2a+(n-1)d)}=\frac{m^{2}}{n^{2}}$
$\displaystyle \frac{m(2a+(m-1)d)}{n(2a+(n-1)d)}=\frac{m^{2}}{n^{2}}$
$\displaystyle \frac{2a+(m-1)d}{2a+(n-1)d}=\frac{m}{n}$
$\displaystyle n(2a+(m-1)d)$ $\displaystyle =m(2a+(n-1)d$
$\displaystyle 2an+mnd-nd$ $\displaystyle =2am+mnd-md$
$\displaystyle 2an-nd=2am-md$
$\displaystyle 2an-2am=-md+nd$
$\displaystyle 2a(n-m)=d(n-m)$
$\displaystyle d=2a$
Now, $\displaystyle \frac{a_{m}}{a_{n}}=\frac{a+(m-1)d}{a+(n-1)d}$
$\displaystyle \frac{a_{m}}{a_{n}}=\frac{a+(m-1)2a}{a+(n-1)2a}$
$\displaystyle =\frac{a+(m-1)2a}{a+(n-1)2a}$
$\displaystyle =\frac{a+2am-2a}{a+2an-2a}$
$\displaystyle =\frac{2am-a}{2an-a}=\frac{a(2m-1)}{a(2n-1)}$
$\displaystyle \frac{a_{m}}{a_{n}}=\frac{2m-1}{2n-1}$
13. If the sum of n terms of an A.P. is $\displaystyle 3n^{2}+5n$ and its $\displaystyle m^{th}$ term is 164, find the value of m.
Given, $\displaystyle S-{n}=3n^{2}+5n$
$\displaystyle a_{m}=164$
We know that,
$\displaystyle a_{n}=S_{n}-S_{n-1}$
$\displaystyle a_{n}=3n^{2}+5n-(3(n-1)^{2}$ $\displaystyle +5(n-1))$
$\displaystyle =3n^{2}+5n-(3(n^{2}+1-2n)$ $\displaystyle +5n-5)$
$\displaystyle =3n^{2}+5n-3n^{2}-3$ $\displaystyle +6n-5n+5$
$\displaystyle a_{n}=6n+2$
Now, $\displaystyle a_{m}=6m+2=164$
$\displaystyle 6m=162$
$\displaystyle m=27$
14. Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.
Let $\displaystyle A_{1}, A_{2},A_{3}, A_{4},A_{5}$ are the five numbers between 8 and 26 such that $\displaystyle 8,A_{1}, A_{2},A_{3}, A_{4},A_{5},26$ are in A.P.
$\displaystyle a=8, l=26, n=7$
We know that, $\displaystyle a_{n}=l=a+(n-1)d$
$\displaystyle 26=8+(7-1)d$
$\displaystyle 18=6d$
$\displaystyle d=3$
Now,
$\displaystyle A_{1}=a+d=8+3=11$
$\displaystyle A_{2}=a+2d=8+6=14$
$\displaystyle A_{3}=a+3d=8+9=17$
$\displaystyle A_{4}=a+4d=8+12=20$
$\displaystyle A_{5}=a+5d=8+15=23$
Hence, five numbers between 8 and 26 are 11,14,17,20,23
15. If $\displaystyle \frac{a^{n}+b^{n}}{a^{n-1}+b^{n-1}}$ is the A.M. between a and b, then find the value of n.
Given $\displaystyle A=\frac{a^{n}+b^{n}}{a^{n-1}+b^{n-1}}$
We know that, $\displaystyle A=\frac{a+b}{2}$
$\displaystyle \frac{a+b}{2}=\frac{a^{n}+b^{n}}{a^{n-1}+b^{n-1}}$
$\displaystyle (a+b)(a^{n-1}+b^{n-1})=2(a^{n}+b^{n})$
$\displaystyle a^{n}+ab^{n-1}+ba^{n-1}+b^{n}$ $\displaystyle =2(a^{n}+b^{n})$
$\displaystyle ab^{n-1}+ba^{n-1}=a^{n}+b^{n}$
$\displaystyle ab^{n-1}-b^{n}=a^{n}-+ba^{n-1}$
$\displaystyle b^{n-1}(a-b)=a^{n-1}(a-b)$
$\displaystyle b^{n-1}=a^{n-1}$
$\displaystyle \frac{a^{n-1}}{b^{n-1}}=1$
$\displaystyle (\frac{a}{b})^{n-1}=(\frac{a}{b})^{0}$
$\displaystyle n-1=0$
$\displaystyle n=1$
16. Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of $\displaystyle 7^{th}$ and $\displaystyle (m-1)^{th}$ numbers is $\displaystyle 5:9$. Find the value of m.
Let $\displaystyle A_{1}, A_{2},A_{3},...A_{m}$ are m numbers between 1 and 31.
Given, $\displaystyle a=1, l=a_{m+2}=31$
We know that,
$\displaystyle l=a+(n-1)d$
$\displaystyle 31=1+(m+2-1)d$
$\displaystyle 30=(m+1)d$
$\displaystyle d=\frac{30}{m+1}$
Given, $\displaystyle \frac{A_{7}}{A_{m-1}}=\frac{5}{9}$
$\displaystyle \frac{a+7d}{a+(m-1)d}=\frac{5}{9}$
$\displaystyle \large \frac{1+7\frac{30}{m+1}}{1+(m-1)\frac{30}{m+1}}=\frac{5}{9}$
$\displaystyle \frac{m+1+210}{m+1+30m-30}=\frac{5}{9}$
$\displaystyle \frac{m+1+210}{31m-29}=\frac{5}{9}$
$\displaystyle 9m+1899=155m-145$
$\displaystyle 146m=2044$
$\displaystyle m=14$
17. A man starts repaying a loan as first installment of Rs.100. If he increases the installment by Rs.5 every month, what amount he will pay in the $\displaystyle 30^{th}$ installment?
Given, first installment=100, second installment=105 and so on...
$\displaystyle a=100, d=105-100=5$
We know that, $\displaystyle a_{n}=a+(n-1)d$
$\displaystyle 30^{th}$ installment is
$\displaystyle a_{30}=100+(30-1)5$
$\displaystyle a_{30}=100+29*5$
$\displaystyle a_{30}=100+145=245$
18. The difference between any two consecutive interior angles of a polygon is $\displaystyle 5^{0}$. If the smallest angle is $\displaystyle 120^{0}$, find the number of the sides of the polygon.
Let the number of sides of the polygon be n.
Given, $\displaystyle a=120, d=5$
We know that, sum of all interior angles of a polygon is $\displaystyle (n-2)180^{0}$
∴ $\displaystyle S_{n}=(n-2)180^{0}$
$\displaystyle \frac{n}{2}(2a+(n-1)d)=(n-2)180$
$\displaystyle \frac{n}{2}(2*120+(n-1)5)=(n-2)180$
$\displaystyle n(240+5n-5)=2(n-2)180$
$\displaystyle n(235+5n)=(n-2)360$
$\displaystyle 235n+5n^{2}=360n-720$
$\displaystyle 5n^{2}-125n+720=0$
$\displaystyle n^{2}-25n+144=0$
$\displaystyle (n-16)(n-9)=0$
$\displaystyle n=16$ or $\displaystyle n=9$
We know that, $\displaystyle a_{n}=a+(n-1)d$
$\displaystyle n=16$ ⇒ $\displaystyle a_{16}=120+15*5=195^{0}>180^{0}$ which is not possible.
$\displaystyle n=9$ ⇒ $\displaystyle a_{9}=120+8*5=160^{0}<180^{0}$
∴ $\displaystyle n=9$
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