1. Find the $\displaystyle 20^{th}$ and $\displaystyle n^{th}$ terms of the G.P. $\displaystyle \frac{5}{2},\frac{5}{4},\frac{5}{8},...$
Given,
$\displaystyle a=\frac{5}{2}$
$\displaystyle r=\frac{5/4}{5/2}=\frac{2}{4}=\frac{1}{2}$
$\displaystyle a_{n}=ar^{n-1}$
$\displaystyle a_{20}=ar^{20-1}=\frac{5}{2}\left ( \frac{1}{2} \right )^{19}$ $\displaystyle =\frac{5}{2}\times \frac{1}{2^{19}}=\frac{5}{2^{20}}$
$\displaystyle a_{n}=ar^{n-1}=\frac{5}{2}\times \left (\frac{1}{2} \right )^{n-1}$ $\displaystyle =\frac{5}{2}\times \frac{1}{2^{n-1}}=\frac{5}{2^{n}}$
2. Find the $\displaystyle 12^{th}$ term of a G.P. whose $\displaystyle 8^{th}$ term is 192 an the common ratio is 2.
Given,
$\displaystyle r=2$
$\displaystyle a_{8}=192$
$\displaystyle ar^{8-1}=192$
$\displaystyle a2^{7}=192$
$\displaystyle a=\frac{192}{128}=\frac{3}{2}$
Therefore,
$\displaystyle a_{12}=ar^{12-1}$ $\displaystyle =\frac{3}{2}\times 2^{11}=3072$
3. The $\displaystyle 5^{th}$, $\displaystyle 8^{th}$ and $\displaystyle 11^{th}$ terms of a G.P. are p, q and s respectively. Show that $\displaystyle q^{2}=ps$.
Given,
$\displaystyle a_{5}=p=ar^{4}$
$\displaystyle a_{8}=q=ar^{7}$
$\displaystyle a_{11}=s=ar^{10}$
$\displaystyle q^{2}=(ar^{7})^{2}=a^{2}r^{14}$ ...(i)
$\displaystyle ps=ar^{4}\times ar^{10}=a^{2}r^{14}$ ...(ii)
from (i) and (ii)
$\displaystyle q^{2}=ps$
4. The $\displaystyle 4^{th}$ term of a G.P. is square of its second term, and the first term is -3. Determine its $\displaystyle 7^{th}$ term.
Given,
a=-3
$\displaystyle a_{4}=(a_{2})^{2}$
$\displaystyle ar^{3}=(ar)^{2}$
$\displaystyle a=r=-3$
$\displaystyle a_{7}=ar^{6}=(-3)(-3)^{6}=-2187$
5. Which term of the following sequences:
a) $\displaystyle 2,2\sqrt{2},4,...$ is 128
Given,
$\displaystyle a=2$
$\displaystyle r=\frac{2\sqrt{2}}{2}=\sqrt{2}$
$\displaystyle a_{n}=ar^{n-1}=128$
$\displaystyle 2(\sqrt{2})^{n-1}=128$
$\displaystyle (\sqrt{2})^{n+1}=(\sqrt{2})^{14}=128$
$\displaystyle n+1=14\Rightarrow n=13$
$\displaystyle 13^{th}$ term is 128
b) $\displaystyle \sqrt{3},3,3\sqrt{2},...$ is 729
Given,
$\displaystyle a=\sqrt{3}$
$\displaystyle r=\frac{3}{\sqrt{3}}=\sqrt{3}$
$\displaystyle a_{n}=ar^{n-1}=729$
$\displaystyle \sqrt{3}(\sqrt{3})^{n-1}=729$
$\displaystyle (\sqrt{3})^{n}=(\sqrt{3})^{12}=729$
$\displaystyle n=12$
$\displaystyle 12^{th}$ term is 729
c) $\displaystyle \frac{1}{3},\frac{1}{9},\frac{1}{27},... is \frac{1}{19683}$
Given,
$\displaystyle a=\frac{1}{3}$
$\displaystyle r=\frac{1/9}{1/3}=\frac{1}{3}$
$\displaystyle a_{n}=ar^{n-1}=\frac{1}{19683}$
$\displaystyle \frac{1}{3}(\frac{1}{3})^{n-1}=729$
$\displaystyle \frac{1}{3^{n}}=\frac{1}{3^{9}}=\frac{1}{19683}$
$\displaystyle n=9$
$\displaystyle 9^{th}$ term is $\displaystyle \frac{1}{19683}$
6. For what values of x, the numbers $\displaystyle -\frac{2}{7},x,-\frac{7}{2}$ are in G.P.?
common ratio, $\displaystyle r=\frac{x}{-2/7}=\frac{-7/2}{x}$
$\displaystyle x^{2}=\frac{-7}{2}\times \frac{-2}{7}=1$
$\displaystyle x=\pm 1$
Find the sum to indicated number of terms in each of the geometric progressions.
7. 0.15, 0.015, 0.0015,... 20 terms.
Given,
$\displaystyle a=0.15$
$\displaystyle r=\frac{0.015}{0.15}=0.1$
$\displaystyle S_{n}=\frac{a(1-r^{n})}{1-r}$ [r<1]
$\displaystyle S_{20}=\frac{0.15[1-(0.1)^{20}]}{1-0.1}$
$\displaystyle S_{20}=\frac{0.15}{0.9}[1-(0.1)^{20}]$
$\displaystyle S_{20}=\frac{1}{6}[1-(0.1)^{20}]$
8. $\displaystyle \sqrt{7},\sqrt{21},3\sqrt{7},...$ n terms
Given,
$\displaystyle a=\sqrt{7}$
$\displaystyle r=\frac{\sqrt{21}}{\sqrt{7}}=\sqrt{3}$
$\displaystyle S_{n}=\frac{a(r^{n}-1)}{r-1}$ [r>1]
$\displaystyle S_{n}=\frac{\sqrt{7}((\sqrt{3})^{n}-1)}{\sqrt{3}-1}$
$\displaystyle S_{n}=\frac{\sqrt{7}((\sqrt{3})^{n}-1)}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}$
$\displaystyle S_{n}=\frac{\sqrt{7}((\sqrt{3})^{n}-1)(\sqrt{3}+1)}{3-1}$
$\displaystyle S_{n}=\frac{\sqrt{7}{2}}((\sqrt{3})^{n}-1)(\sqrt{3}+1)$
9. $\displaystyle 1,-a,a^{2},-a^{3},...$ n terms if ($\displaystyle a\neq -1$)
Given,
$\displaystyle A=1$
$\displaystyle R=\frac{-a}{1}=-a$
$\displaystyle S_{n}=\frac{a(1-R^{n})}{1-R}$ [r<1]
$\displaystyle S_{n}=\frac{1[1-(-a)^{n}]}{1-(-a)}$
$\displaystyle S_{n}=\frac{[1-(-a)^{n}]}{1+a}$
10. $\displaystyle x^{3},x^{5},x^{7},...$ n terms if ($\displaystyle x\neq \pm 1$)
Given,
$\displaystyle a=x^{3}$
$\displaystyle r=\frac{x^{5}}{x^{3}}=x^{2}$
$\displaystyle S_{n}=\frac{a(1-r^{n})}{1-r}$
$\displaystyle S_{n}=\frac{x^{3}(1-(x^{2})^{n})}{1-x^{2}}$
$\displaystyle S_{n}=\frac{x^{3}(1-x^{2n})}{1-x^{2}}$
11. Evaluate $\displaystyle \sum_{k=1}^{11}(2+3^{k})$
$\displaystyle \sum_{k=1}^{11}(2+3^{k})$ $\displaystyle =\sum_{k=1}^{11}2+\sum_{k=1}^{11}3^{k}$
$\displaystyle =22+3+3^{2}+3^{3}+...+3^{11}$
From G.P. $\displaystyle 3,3^{2},3^{3},...3^{11}$
$\displaystyle a=3$
$\displaystyle r=\frac{3^{2}}{3}=3$
$\displaystyle S_{n}=\frac{a(r^{n}-1)}{r-1}$ (r>1)
$\displaystyle S_{11}=\frac{3(3^{11}-1)}{3-1}$
$\displaystyle S_{11}=\frac{3}{2}(3^{11}-1)$
Therefore, $\displaystyle 22+\frac{3}{2}(3^{11}-1)$
12. The sum of first three terms of a G.P. is $\displaystyle \frac{39}{10}$ and their product is 1. Find the common ratio and the terms.
Let $\displaystyle \frac{a}{r},a,ar$ be the terms.
Given,
$\displaystyle \frac{a}{r}\times a\times ar=1$ $\displaystyle \Rightarrow a^{3}=1$ $\displaystyle \Rightarrow a=1$
$\displaystyle \frac{a}{r}+a+ar=\frac{39}{10}$ $\displaystyle \Rightarrow \frac{a+ar+ar^{2}}{r}=\frac{39}{10}$ $\displaystyle \Rightarrow 1+r+r^{2}=\frac{39}{10}r$ $\displaystyle \Rightarrow 10r^{2}-29r+10=0$
$\displaystyle r=\frac{5}{2}$ or $\displaystyle r=\frac{2}{5}$
∴ the terms are $\displaystyle \frac{5}{2},1,\frac{2}{5}$
13. How many terms of G.P. $\displaystyle 3,3^{2},3^{3},...$ are needed to give the sum 120?
Given,
$\displaystyle a=3$
$\displaystyle r=\frac{3^{2}}{3}=3$
$\displaystyle S_{n}=\frac{a(r^{n}-1)}{r-1}$ [r>1]
$\displaystyle 120=\frac{3(3^{n}-1)}{3-1}$
$\displaystyle 120=\frac{3(3^{n}-1)}{2}$
$\displaystyle 80=3^{n}-1$
$\displaystyle 81=3^{n}$
$\displaystyle 81=3^{4}$
$\displaystyle n=4$
14. The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.
Given,
$\displaystyle a+ar+ar^{2}=16$ $\displaystyle \Rightarrow a(1+r+r^{2})=16$ ...(i)
$\displaystyle ar^{3}+ar^{4}+ar^{5}=128$ $\displaystyle \Rightarrow ar^{3}(1+r+r^{2})=128$ ...(ii)
Divide (ii) by (i)
$\displaystyle \frac{ar^{3}(1+r+r^{2})}{a(1+r+r^{2})}=\frac{128}{16}$
$\displaystyle r^{3}=8=2^{3}$ $\displaystyle \Rightarrow r=2$
$\displaystyle (i)\Rightarrow a(1+2+2^{2})=16$
$\displaystyle a=\frac{16}{7}$
$\displaystyle S_{n}=\frac{a(r^{n}-1)}{r-1}$ [r>1]
$\displaystyle S_{n}=\frac{\frac{16}{7}(2^{n}-1)}{2-1}$
$\displaystyle S_{n}=\frac{16}{7}(2^{n}-1)$
15. Given a G.P. with a=729 and $\displaystyle 7^{th}$ term 64, determine $\displaystyle S_{7}$.
Given,
$\displaystyle a=729$
$\displaystyle a_{7}=ar^{6}=64$
$\displaystyle r^{6}=\frac{64}{729}$ $\displaystyle \Rightarrow r^{6}=\left (\frac{2}{3} \right )^{6}$
$\displaystyle r=\frac{2}{3}$
$\displaystyle S_{n}=\frac{a(1-r^{n})}{1-r}$ [r<1]
$\displaystyle S_{7}=\frac{729(1-(\frac{2}{3})^{7})}{1-(\frac{2}{3})}$
$\displaystyle S_{7}=\frac{729(1-\frac{128}{2187})}{\frac{1}{3}}$
$\displaystyle \Rightarrow S_{7}=2059$
16. Find a G.P. for which sum of the first two terms is -4 and the fifth term is 4 times the third term.
Given,
$\displaystyle a_{5}=4a_{3}$ $\displaystyle \Rightarrow ar^{4}=4ar^{2}$ $\displaystyle \Rightarrow r^{2}=4$
$\displaystyle r=\pm 2$
$\displaystyle a+ar=-4$ $\displaystyle \Rightarrow a(1+r)=-4$
when r=2, $\displaystyle a=\frac{-4}{3}$
when r=-2, $\displaystyle a=4$
∴ G.P. is $\displaystyle -\frac{4}{3},-\frac{8}{3},-\frac{16}{3}...$ or $\displaystyle 4,-8,16,...$
17. If the $\displaystyle 4^{th}$, $\displaystyle 10^{th}$ and $\displaystyle 16^{th}$ terms of a G.P. are x, y and z respectively. Prove that x, y, z are in G.P.
Given,
$\displaystyle a_{4}=ar^{3}=x$
$\displaystyle a_{10}=ar^{9}=y$
$\displaystyle a_{16}=ar^{15}=x$
divide y by x,
$\displaystyle \frac{y}{x}=\frac{ar^{9}}{ar^{3}}=r^{6}$
divide z by y,
$\displaystyle \frac{z}{y}=\frac{ar^{15}}{ar^{9}}=r^{6}$
Since $\displaystyle \frac{y}{x}=\frac{z}{y}$, x, y, z are in G.P.
18. Find the sum to n terms of the sequence, 8, 88, 888, 8888,...
$\displaystyle S_{n}=8+8+888+8888+...$
$\displaystyle S_{n}=8[1+11+111+1111+...]$
$\displaystyle S_{n}=\frac{8}{9}[9+99+999+9999+...]$
$\displaystyle S_{n}=\frac{8}{9}[(10-1)+(10^{2}-1)$ $\displaystyle +(10^{3}-1)+...]$
$\displaystyle S_{n}=\frac{8}{9}[10+10^{2}+10^{3}+...$ $\displaystyle -(1+1+1+...]$
$\displaystyle S_{n}=\frac{8}{9}[\frac{10(10^{n}-1)}{10-1}+...-n]$
$\displaystyle S_{n}=\frac{8}{9}[\frac{10}{9}(10^{n}-1)+...-n]$
$\displaystyle S_{n}=\frac{80}{81}(10^{n}-1)-\frac{8}{9}n$
19. Find the sum of the products of the corresponding terms of the sequences 2,4,8,16,32 and 128,32,8,2,1/2.
Sum=$\displaystyle 2\times 128+4\times 32+8\times 8$ $\displaystyle +16\times 2+32\times \frac{1}{2}$
$\displaystyle =64[4+2+2+\frac{1}{2}+\frac{1}{2^{2}}]$
$\displaystyle a=4$
$\displaystyle r=\frac{1}{2}$
$\displaystyle S_{n}=\frac{a(1-r^{n})}{1-r}$ [r<1]
$\displaystyle S_{5}=\frac{4(1-(\frac{1}{2})^{5})}{1-(\frac{1}{2})}$
$\displaystyle S_{5}=\frac{4(1-\frac{1}{32}}{\frac{1}{2}}$
$\displaystyle S_{5}=\frac{31}{4}$
Sum=$\displaystyle 64\times \frac{31}{4}=496$
20. Show that the products of the corresponding terms of the sequences $\displaystyle a,ar,ar^{2},...ar^{n-1}$ and $\displaystyle A,AR,AR^{2},...AR^{n-1}$ form a G.P. and find the common ratio.
Product of the sequence is $\displaystyle aA,arAR,ar^{2}AR^{2},...$
$\displaystyle \frac{2^{nd}term}{1^{st}term}=\frac{arAR}{aA}=rR$
$\displaystyle \frac{3^{rd}term}{2^{nd}term}=\frac{ar^{2}AR^{2}}{arAR}=rR$
Since $\displaystyle \frac{2^{nd}term}{1^{st}term}=\frac{3^{rd}term}{2^{nd}term}$ the sequence $\displaystyle aA,arAR,ar^{2}AR^{2},...$ is a G.P.
21. Find four numbers forming a geometric progression in which the third term is greater than the first term by 9 and the second term is greater than the $\displaystyle 4^{th}$ by 18.
Given,
$\displaystyle a_{3}=a+9$ $\displaystyle \Rightarrow ar^{2}=a+9$ $\displaystyle \Rightarrow a(r^{2}-1)=9$ ...(i)
$\displaystyle a_{2}=a_{4}+18$ $\displaystyle \Rightarrow ar=ar^{3}+18$ $\displaystyle \Rightarrow ar(1-r^{2})=18$ ...(ii)
divide (ii) by (i),
$\displaystyle \frac{ar(1-r^{2})}{a(r^{2}-1)}=\frac{18}{9}$
$\displaystyle r=-2$
$\displaystyle (i) \Rightarrow a((-2)^{2}-1)=9$
$\displaystyle 3a=9$
$\displaystyle a=3$
The terms of G.P. are 3, -6, 12, -24
22. If the $\displaystyle p^{th},q^{th},r^{th}$ terms of a G.P. are a, b and c respectively. Prove that $\displaystyle a^{q-r}b^{r-p}c^{p-q}=1$
Given,
$\displaystyle a_{p}=a=AR^{p-1}$
$\displaystyle a_{q}=b=AR^{q-1}$
$\displaystyle a_{r}=a=AR^{r-1}$
$\displaystyle a^{q-r}b^{r-p}c^{p-q}$ $=\displaystyle A^{q-r}R^{(p-1)(q-r)}B^{r-p}R^{(q-1)(r-p)}$ $\displaystyle C^{p-q}R^{(r-1)(p-q)}$
$\displaystyle =A^{q-r+r-p+p-q}$ $\displaystyle R^{pq-pr-q+r+qr-pq-r+p+pr+q-p-qr}$
$\displaystyle =A^{0}R^{0}=1$
23. If the first and the $\displaystyle n^{th}$ term of a G.P. are a and b respectively and if P is the product of n terms, prove that $\displaystyle P^{2}=(ab)^{n}$
Given,
first term is a, last term is b.
G.P. is $\displaystyle a,ar,ar^{2},...ar^{n-1}$
product $\displaystyle P=(a)(ar)(ar^{2})...(ar^{n-1})$
$\displaystyle =(a\times a\times ...a)(r\times r^{2}\times r^{3}...r^{n-1})$
$\displaystyle =(a^{n})(r^{1+2+...(n-1)})$
$\displaystyle =(a^{n})(r^{\frac{n(n-1)}{2}})$
$\displaystyle P^{2}=(a^{2n})(r^{n(n-1)})$
$\displaystyle P^{2}=[a^{2}r^{n-1}]^{n}$
$\displaystyle P^{2}=[a.ar^{n-1}]^{n}$
$\displaystyle P^{2}=[a.b]^{n}$
24. Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from $\displaystyle (n+1)^{th}$ to $\displaystyle (2n)^{th}$ term is $\displaystyle \frac{1}{r^{n}}$.
Given,
$\displaystyle S_{n}=\frac{a(1-r^{n})}{1-r}$ ...(i)
Since there are n terms from (n+1) to 2n,
$\displaystyle S_{n}=\frac{a_{n+1}(1-r^{n})}{1-r}$ $\displaystyle \Rightarrow S_{n}=\frac{ar^{n}(1-r^{n})}{1-r}$ ...(ii)
Required ratio is,
$\displaystyle \frac{(i)}{(ii)}$ $\displaystyle \Rightarrow \frac{\frac{a(1-r^{n})}{1-r}}{\frac{ar^{n}(1-r^{n})}{1-r}}$
$\displaystyle \frac{a(1-r^{n})}{1-r}\times \frac{1-r}{ar^{n}(1-r^{n})}$
$\displaystyle =\frac{1}{r^{n}}$
26. Intsert two numbers between 3 and 81 so that the resulting sequence is GP.
Let $\displaystyle a=3$
$\displaystyle a_{1},a_{2}$ are two numbers between 3 and 81
$\displaystyle a_{4}=81$ $\displaystyle \Rightarrow ar^{3}=81$
$\displaystyle 3r^{3}=81$
$\displaystyle r^{3}=27$ $\displaystyle \Rightarrow 3^{3}=27$
$\displaystyle r=3$
Therefore,
$\displaystyle a_{2}=ar=3\times 3=9$
$\displaystyle a_{3}=ar^{2}=3\times 3^{2}=27$
Two numbers between 3 and 81 are 9 and 27.
27. Find the value of n so that $\displaystyle \frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}$ may be the geometric mean between a and b.
We know that geometric mean between a and b is given by $\displaystyle \sqrt{ab}$
$\displaystyle \frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}=\sqrt{ab}$
squaring on both sides,
$\displaystyle \left (\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}} \right )^{2}=ab$
$\displaystyle \frac{a^{2(n+1)}+b^{2(n+1)}+2a^{n+1}b^{n+1}}{a^{2n}+b^{2n}+2a^{n}b^{n}}=ab$
$\displaystyle a^{2n+2}+b^{2n+2}+2a^{n+1}b^{n+1}$ $\displaystyle =ab(a^{2n}+b^{2n}+2a^{n}b^{n})$
$\displaystyle a^{2n+2}+b^{2n+2}+2a^{n+1}b^{n+1}$ $\displaystyle =a^{2n+1}b+ab^{2n+1}+2a^{n+1}b^{n+1}$
$\displaystyle a^{2n+2}+b^{2n+2}=a^{2n+1}b+ab^{2n+1}$
$\displaystyle a^{2n+2}-a^{2n+1}b=ab^{2n+1}-b^{2n+2}$
$\displaystyle a^{2n+1}(a-b)=b^{2n+1}(a-b)$
$\displaystyle a^{2n+1}=b^{2n+1}$
$\displaystyle \left ( \frac{a}{b} \right )^{2n+1}=1=\left ( \frac{a}{b} \right )^{0}$
$\displaystyle \Rightarrow 2n+1=0$
$\displaystyle n=-\frac{1}{2}$
28. The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio $\displaystyle (3+2\sqrt{2}):(3-2\sqrt{2})$.
Given,
$\displaystyle a+b=6\sqrt{ab}$ ...(i)
squaring on both sides,
$\displaystyle (a+b)^{2}=36ab$
we know that,
$\displaystyle (a-b)^{2}=(a+b)^{2}-4ab$
$\displaystyle (a-b)^{2}=36ab-4ab$
$\displaystyle (a-b)^{2}=32ab$
$\displaystyle a-b=\sqrt{32}\sqrt{ab}$
$\displaystyle a-b=4\sqrt{2}\sqrt{ab}$ ...(ii)
adding (i) and (ii)
$\displaystyle a+b+a-b=6\sqrt{ab}+4\sqrt{2}\sqrt{ab}$
$\displaystyle 2a=(6+4\sqrt{2})\sqrt{ab}$
$\displaystyle 2a=2(3+2\sqrt{2})\sqrt{ab}$
$\displaystyle a=(3+2\sqrt{2})\sqrt{ab}$
$\displaystyle (i)\Rightarrow b=6\sqrt{ab}-a$
$\displaystyle b=6\sqrt{ab}-(3+2\sqrt{2})\sqrt{ab}$
$\displaystyle b=(6-3-2\sqrt{2})\sqrt{ab}$
$\displaystyle b=(3-2\sqrt{2})\sqrt{ab}$
$\displaystyle \frac{a}{b}=\frac{(3+2\sqrt{2})\sqrt{ab}}{(3-2\sqrt{2})\sqrt{ab}}$ $\displaystyle =\frac{3+2\sqrt{2}}{3-2\sqrt{2}}$
Thus, $\displaystyle a:b=(3+2\sqrt{2}):(3-2\sqrt{2})$
29. If A and G be AM and GM respectively between two positive numbers, prove that the numbers are $\displaystyle A\pm \sqrt{(A+G)(A-G)}$
Let the two numbers be a and b
given that,
$\displaystyle AM=A=\frac{a+b}{2}$ $\displaystyle \Rightarrow a+b=2A$ ...(i)
$\displaystyle GM=G=\sqrt{ab}$ $\displaystyle \Rightarrow ab=G^{2}$
We know that,
$\displaystyle (a-b)^{2}=(a+b)^{2}-4ab$
$\displaystyle (a-b)^{2}=(2A)^{2}-4G^{2}$
$\displaystyle (a-b)^{2}=4A^{2}-4G^{2}$
$\displaystyle (a-b)^{2}=4(A^{2}-G^{2})$
$\displaystyle a-b=\pm \sqrt{4(A^{2}-G^{2})}$
$\displaystyle a-b=\pm 2\sqrt{(A-G)(A+G)}$
$\displaystyle a-(2A-a)$ $\displaystyle =\pm 2\sqrt{(A-G)(A+G)}$ [$\displaystyle b=2A-a$ ]
$\displaystyle a-2A+a=\pm 2\sqrt{(A-G)(A+G)}$
$\displaystyle 2a-2A=\pm 2\sqrt{(A-G)(A+G)}$
$\displaystyle 2a=2A\pm 2\sqrt{(A-G)(A+G)}$
$\displaystyle a=A\pm \sqrt{(A-G)(A+G)}$
$\displaystyle b=2A-A\pm \sqrt{(A-G)(A+G)}$
$\displaystyle b=A\pm \sqrt{(A-G)(A+G)}$
30. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of $\displaystyle 2^{nd}$ hour, $\displaystyle 4^{th}$ hour and $\displaystyle n^{th}$ hour?
Given that initial number of bacteria is 30 and doubles every hour.
∴ a=30, r=2
Number of bacteria at the end of $\displaystyle 2^{nd}$ hour, $\displaystyle a_{3}=ar^{2}=30(2)^{2}=120$
Number of bacteria at the end of $\displaystyle 4^{th}$ hour, $\displaystyle a_{5}=ar^{4}=30(2)^{4}=480$
Number of bacteria at the end of $\displaystyle n^{th}$ hour, $\displaystyle a_{n+1}=ar^{n}=30(2)^{n}$
31. What will Rs 500 amounts to 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?
Given that,
P = 500
r = 10
We know that, amount for compound interest, $\displaystyle A=P\left ( 1+\frac{r}{100} \right )^{n}$
amount at the end of first year = $\displaystyle 500\left ( 1+\frac{10}{100} \right )^{1}$ = $\displaystyle 500\left ( 1+\frac{1}{10} \right )$ = $\displaystyle 500\left ( \frac{11}{10} \right )$ = $\displaystyle 500(1.1)$
amount at the end of second year = $\displaystyle 500(1.1)(1.1)=500(1.1)^{2}$
amount at the end of third year = $\displaystyle 500(1.1)(1.1)(1.1)=500(1.1)^{3}$ so on..
amount at the end of 10 year = $\displaystyle 500(1.1)^{10}$
32. If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.
Let the roots of quadratic equation be a and b
Given that,
$\displaystyle AM=\frac{a+b}{2}=8 \Rightarrow a+b=16$ ...(i)
$\displaystyle GM=\sqrt{ab}=5 \Rightarrow ab=25$ ...(ii)
The quadratic equation is given by,
$\displaystyle x^{2}+(a+b)x+ab=0$
$\displaystyle x^{2}+16x+25=0$
Given,
$\displaystyle a=\frac{5}{2}$
$\displaystyle r=\frac{5/4}{5/2}=\frac{2}{4}=\frac{1}{2}$
$\displaystyle a_{n}=ar^{n-1}$
$\displaystyle a_{20}=ar^{20-1}=\frac{5}{2}\left ( \frac{1}{2} \right )^{19}$ $\displaystyle =\frac{5}{2}\times \frac{1}{2^{19}}=\frac{5}{2^{20}}$
$\displaystyle a_{n}=ar^{n-1}=\frac{5}{2}\times \left (\frac{1}{2} \right )^{n-1}$ $\displaystyle =\frac{5}{2}\times \frac{1}{2^{n-1}}=\frac{5}{2^{n}}$
2. Find the $\displaystyle 12^{th}$ term of a G.P. whose $\displaystyle 8^{th}$ term is 192 an the common ratio is 2.
Given,
$\displaystyle r=2$
$\displaystyle a_{8}=192$
$\displaystyle ar^{8-1}=192$
$\displaystyle a2^{7}=192$
$\displaystyle a=\frac{192}{128}=\frac{3}{2}$
Therefore,
$\displaystyle a_{12}=ar^{12-1}$ $\displaystyle =\frac{3}{2}\times 2^{11}=3072$
3. The $\displaystyle 5^{th}$, $\displaystyle 8^{th}$ and $\displaystyle 11^{th}$ terms of a G.P. are p, q and s respectively. Show that $\displaystyle q^{2}=ps$.
Given,
$\displaystyle a_{5}=p=ar^{4}$
$\displaystyle a_{8}=q=ar^{7}$
$\displaystyle a_{11}=s=ar^{10}$
$\displaystyle q^{2}=(ar^{7})^{2}=a^{2}r^{14}$ ...(i)
$\displaystyle ps=ar^{4}\times ar^{10}=a^{2}r^{14}$ ...(ii)
from (i) and (ii)
$\displaystyle q^{2}=ps$
4. The $\displaystyle 4^{th}$ term of a G.P. is square of its second term, and the first term is -3. Determine its $\displaystyle 7^{th}$ term.
Given,
a=-3
$\displaystyle a_{4}=(a_{2})^{2}$
$\displaystyle ar^{3}=(ar)^{2}$
$\displaystyle a=r=-3$
$\displaystyle a_{7}=ar^{6}=(-3)(-3)^{6}=-2187$
5. Which term of the following sequences:
a) $\displaystyle 2,2\sqrt{2},4,...$ is 128
Given,
$\displaystyle a=2$
$\displaystyle r=\frac{2\sqrt{2}}{2}=\sqrt{2}$
$\displaystyle a_{n}=ar^{n-1}=128$
$\displaystyle 2(\sqrt{2})^{n-1}=128$
$\displaystyle (\sqrt{2})^{n+1}=(\sqrt{2})^{14}=128$
$\displaystyle n+1=14\Rightarrow n=13$
$\displaystyle 13^{th}$ term is 128
b) $\displaystyle \sqrt{3},3,3\sqrt{2},...$ is 729
Given,
$\displaystyle a=\sqrt{3}$
$\displaystyle r=\frac{3}{\sqrt{3}}=\sqrt{3}$
$\displaystyle a_{n}=ar^{n-1}=729$
$\displaystyle \sqrt{3}(\sqrt{3})^{n-1}=729$
$\displaystyle (\sqrt{3})^{n}=(\sqrt{3})^{12}=729$
$\displaystyle n=12$
$\displaystyle 12^{th}$ term is 729
c) $\displaystyle \frac{1}{3},\frac{1}{9},\frac{1}{27},... is \frac{1}{19683}$
Given,
$\displaystyle a=\frac{1}{3}$
$\displaystyle r=\frac{1/9}{1/3}=\frac{1}{3}$
$\displaystyle a_{n}=ar^{n-1}=\frac{1}{19683}$
$\displaystyle \frac{1}{3}(\frac{1}{3})^{n-1}=729$
$\displaystyle \frac{1}{3^{n}}=\frac{1}{3^{9}}=\frac{1}{19683}$
$\displaystyle n=9$
$\displaystyle 9^{th}$ term is $\displaystyle \frac{1}{19683}$
6. For what values of x, the numbers $\displaystyle -\frac{2}{7},x,-\frac{7}{2}$ are in G.P.?
common ratio, $\displaystyle r=\frac{x}{-2/7}=\frac{-7/2}{x}$
$\displaystyle x^{2}=\frac{-7}{2}\times \frac{-2}{7}=1$
$\displaystyle x=\pm 1$
Find the sum to indicated number of terms in each of the geometric progressions.
7. 0.15, 0.015, 0.0015,... 20 terms.
Given,
$\displaystyle a=0.15$
$\displaystyle r=\frac{0.015}{0.15}=0.1$
$\displaystyle S_{n}=\frac{a(1-r^{n})}{1-r}$ [r<1]
$\displaystyle S_{20}=\frac{0.15[1-(0.1)^{20}]}{1-0.1}$
$\displaystyle S_{20}=\frac{0.15}{0.9}[1-(0.1)^{20}]$
$\displaystyle S_{20}=\frac{1}{6}[1-(0.1)^{20}]$
8. $\displaystyle \sqrt{7},\sqrt{21},3\sqrt{7},...$ n terms
Given,
$\displaystyle a=\sqrt{7}$
$\displaystyle r=\frac{\sqrt{21}}{\sqrt{7}}=\sqrt{3}$
$\displaystyle S_{n}=\frac{a(r^{n}-1)}{r-1}$ [r>1]
$\displaystyle S_{n}=\frac{\sqrt{7}((\sqrt{3})^{n}-1)}{\sqrt{3}-1}$
$\displaystyle S_{n}=\frac{\sqrt{7}((\sqrt{3})^{n}-1)}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}$
$\displaystyle S_{n}=\frac{\sqrt{7}((\sqrt{3})^{n}-1)(\sqrt{3}+1)}{3-1}$
$\displaystyle S_{n}=\frac{\sqrt{7}{2}}((\sqrt{3})^{n}-1)(\sqrt{3}+1)$
9. $\displaystyle 1,-a,a^{2},-a^{3},...$ n terms if ($\displaystyle a\neq -1$)
Given,
$\displaystyle A=1$
$\displaystyle R=\frac{-a}{1}=-a$
$\displaystyle S_{n}=\frac{a(1-R^{n})}{1-R}$ [r<1]
$\displaystyle S_{n}=\frac{1[1-(-a)^{n}]}{1-(-a)}$
$\displaystyle S_{n}=\frac{[1-(-a)^{n}]}{1+a}$
10. $\displaystyle x^{3},x^{5},x^{7},...$ n terms if ($\displaystyle x\neq \pm 1$)
Given,
$\displaystyle a=x^{3}$
$\displaystyle r=\frac{x^{5}}{x^{3}}=x^{2}$
$\displaystyle S_{n}=\frac{a(1-r^{n})}{1-r}$
$\displaystyle S_{n}=\frac{x^{3}(1-(x^{2})^{n})}{1-x^{2}}$
$\displaystyle S_{n}=\frac{x^{3}(1-x^{2n})}{1-x^{2}}$
11. Evaluate $\displaystyle \sum_{k=1}^{11}(2+3^{k})$
$\displaystyle \sum_{k=1}^{11}(2+3^{k})$ $\displaystyle =\sum_{k=1}^{11}2+\sum_{k=1}^{11}3^{k}$
$\displaystyle =22+3+3^{2}+3^{3}+...+3^{11}$
From G.P. $\displaystyle 3,3^{2},3^{3},...3^{11}$
$\displaystyle a=3$
$\displaystyle r=\frac{3^{2}}{3}=3$
$\displaystyle S_{n}=\frac{a(r^{n}-1)}{r-1}$ (r>1)
$\displaystyle S_{11}=\frac{3(3^{11}-1)}{3-1}$
$\displaystyle S_{11}=\frac{3}{2}(3^{11}-1)$
Therefore, $\displaystyle 22+\frac{3}{2}(3^{11}-1)$
12. The sum of first three terms of a G.P. is $\displaystyle \frac{39}{10}$ and their product is 1. Find the common ratio and the terms.
Let $\displaystyle \frac{a}{r},a,ar$ be the terms.
Given,
$\displaystyle \frac{a}{r}\times a\times ar=1$ $\displaystyle \Rightarrow a^{3}=1$ $\displaystyle \Rightarrow a=1$
$\displaystyle \frac{a}{r}+a+ar=\frac{39}{10}$ $\displaystyle \Rightarrow \frac{a+ar+ar^{2}}{r}=\frac{39}{10}$ $\displaystyle \Rightarrow 1+r+r^{2}=\frac{39}{10}r$ $\displaystyle \Rightarrow 10r^{2}-29r+10=0$
$\displaystyle r=\frac{5}{2}$ or $\displaystyle r=\frac{2}{5}$
∴ the terms are $\displaystyle \frac{5}{2},1,\frac{2}{5}$
13. How many terms of G.P. $\displaystyle 3,3^{2},3^{3},...$ are needed to give the sum 120?
Given,
$\displaystyle a=3$
$\displaystyle r=\frac{3^{2}}{3}=3$
$\displaystyle S_{n}=\frac{a(r^{n}-1)}{r-1}$ [r>1]
$\displaystyle 120=\frac{3(3^{n}-1)}{3-1}$
$\displaystyle 120=\frac{3(3^{n}-1)}{2}$
$\displaystyle 80=3^{n}-1$
$\displaystyle 81=3^{n}$
$\displaystyle 81=3^{4}$
$\displaystyle n=4$
14. The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.
Given,
$\displaystyle a+ar+ar^{2}=16$ $\displaystyle \Rightarrow a(1+r+r^{2})=16$ ...(i)
$\displaystyle ar^{3}+ar^{4}+ar^{5}=128$ $\displaystyle \Rightarrow ar^{3}(1+r+r^{2})=128$ ...(ii)
Divide (ii) by (i)
$\displaystyle \frac{ar^{3}(1+r+r^{2})}{a(1+r+r^{2})}=\frac{128}{16}$
$\displaystyle r^{3}=8=2^{3}$ $\displaystyle \Rightarrow r=2$
$\displaystyle (i)\Rightarrow a(1+2+2^{2})=16$
$\displaystyle a=\frac{16}{7}$
$\displaystyle S_{n}=\frac{a(r^{n}-1)}{r-1}$ [r>1]
$\displaystyle S_{n}=\frac{\frac{16}{7}(2^{n}-1)}{2-1}$
$\displaystyle S_{n}=\frac{16}{7}(2^{n}-1)$
15. Given a G.P. with a=729 and $\displaystyle 7^{th}$ term 64, determine $\displaystyle S_{7}$.
Given,
$\displaystyle a=729$
$\displaystyle a_{7}=ar^{6}=64$
$\displaystyle r^{6}=\frac{64}{729}$ $\displaystyle \Rightarrow r^{6}=\left (\frac{2}{3} \right )^{6}$
$\displaystyle r=\frac{2}{3}$
$\displaystyle S_{n}=\frac{a(1-r^{n})}{1-r}$ [r<1]
$\displaystyle S_{7}=\frac{729(1-(\frac{2}{3})^{7})}{1-(\frac{2}{3})}$
$\displaystyle S_{7}=\frac{729(1-\frac{128}{2187})}{\frac{1}{3}}$
$\displaystyle \Rightarrow S_{7}=2059$
16. Find a G.P. for which sum of the first two terms is -4 and the fifth term is 4 times the third term.
Given,
$\displaystyle a_{5}=4a_{3}$ $\displaystyle \Rightarrow ar^{4}=4ar^{2}$ $\displaystyle \Rightarrow r^{2}=4$
$\displaystyle r=\pm 2$
$\displaystyle a+ar=-4$ $\displaystyle \Rightarrow a(1+r)=-4$
when r=2, $\displaystyle a=\frac{-4}{3}$
when r=-2, $\displaystyle a=4$
∴ G.P. is $\displaystyle -\frac{4}{3},-\frac{8}{3},-\frac{16}{3}...$ or $\displaystyle 4,-8,16,...$
17. If the $\displaystyle 4^{th}$, $\displaystyle 10^{th}$ and $\displaystyle 16^{th}$ terms of a G.P. are x, y and z respectively. Prove that x, y, z are in G.P.
Given,
$\displaystyle a_{4}=ar^{3}=x$
$\displaystyle a_{10}=ar^{9}=y$
$\displaystyle a_{16}=ar^{15}=x$
divide y by x,
$\displaystyle \frac{y}{x}=\frac{ar^{9}}{ar^{3}}=r^{6}$
divide z by y,
$\displaystyle \frac{z}{y}=\frac{ar^{15}}{ar^{9}}=r^{6}$
Since $\displaystyle \frac{y}{x}=\frac{z}{y}$, x, y, z are in G.P.
18. Find the sum to n terms of the sequence, 8, 88, 888, 8888,...
$\displaystyle S_{n}=8+8+888+8888+...$
$\displaystyle S_{n}=8[1+11+111+1111+...]$
$\displaystyle S_{n}=\frac{8}{9}[9+99+999+9999+...]$
$\displaystyle S_{n}=\frac{8}{9}[(10-1)+(10^{2}-1)$ $\displaystyle +(10^{3}-1)+...]$
$\displaystyle S_{n}=\frac{8}{9}[10+10^{2}+10^{3}+...$ $\displaystyle -(1+1+1+...]$
$\displaystyle S_{n}=\frac{8}{9}[\frac{10(10^{n}-1)}{10-1}+...-n]$
$\displaystyle S_{n}=\frac{8}{9}[\frac{10}{9}(10^{n}-1)+...-n]$
$\displaystyle S_{n}=\frac{80}{81}(10^{n}-1)-\frac{8}{9}n$
19. Find the sum of the products of the corresponding terms of the sequences 2,4,8,16,32 and 128,32,8,2,1/2.
Sum=$\displaystyle 2\times 128+4\times 32+8\times 8$ $\displaystyle +16\times 2+32\times \frac{1}{2}$
$\displaystyle =64[4+2+2+\frac{1}{2}+\frac{1}{2^{2}}]$
$\displaystyle a=4$
$\displaystyle r=\frac{1}{2}$
$\displaystyle S_{n}=\frac{a(1-r^{n})}{1-r}$ [r<1]
$\displaystyle S_{5}=\frac{4(1-(\frac{1}{2})^{5})}{1-(\frac{1}{2})}$
$\displaystyle S_{5}=\frac{4(1-\frac{1}{32}}{\frac{1}{2}}$
$\displaystyle S_{5}=\frac{31}{4}$
Sum=$\displaystyle 64\times \frac{31}{4}=496$
20. Show that the products of the corresponding terms of the sequences $\displaystyle a,ar,ar^{2},...ar^{n-1}$ and $\displaystyle A,AR,AR^{2},...AR^{n-1}$ form a G.P. and find the common ratio.
Product of the sequence is $\displaystyle aA,arAR,ar^{2}AR^{2},...$
$\displaystyle \frac{2^{nd}term}{1^{st}term}=\frac{arAR}{aA}=rR$
$\displaystyle \frac{3^{rd}term}{2^{nd}term}=\frac{ar^{2}AR^{2}}{arAR}=rR$
Since $\displaystyle \frac{2^{nd}term}{1^{st}term}=\frac{3^{rd}term}{2^{nd}term}$ the sequence $\displaystyle aA,arAR,ar^{2}AR^{2},...$ is a G.P.
21. Find four numbers forming a geometric progression in which the third term is greater than the first term by 9 and the second term is greater than the $\displaystyle 4^{th}$ by 18.
Given,
$\displaystyle a_{3}=a+9$ $\displaystyle \Rightarrow ar^{2}=a+9$ $\displaystyle \Rightarrow a(r^{2}-1)=9$ ...(i)
$\displaystyle a_{2}=a_{4}+18$ $\displaystyle \Rightarrow ar=ar^{3}+18$ $\displaystyle \Rightarrow ar(1-r^{2})=18$ ...(ii)
divide (ii) by (i),
$\displaystyle \frac{ar(1-r^{2})}{a(r^{2}-1)}=\frac{18}{9}$
$\displaystyle r=-2$
$\displaystyle (i) \Rightarrow a((-2)^{2}-1)=9$
$\displaystyle 3a=9$
$\displaystyle a=3$
The terms of G.P. are 3, -6, 12, -24
22. If the $\displaystyle p^{th},q^{th},r^{th}$ terms of a G.P. are a, b and c respectively. Prove that $\displaystyle a^{q-r}b^{r-p}c^{p-q}=1$
Given,
$\displaystyle a_{p}=a=AR^{p-1}$
$\displaystyle a_{q}=b=AR^{q-1}$
$\displaystyle a_{r}=a=AR^{r-1}$
$\displaystyle a^{q-r}b^{r-p}c^{p-q}$ $=\displaystyle A^{q-r}R^{(p-1)(q-r)}B^{r-p}R^{(q-1)(r-p)}$ $\displaystyle C^{p-q}R^{(r-1)(p-q)}$
$\displaystyle =A^{q-r+r-p+p-q}$ $\displaystyle R^{pq-pr-q+r+qr-pq-r+p+pr+q-p-qr}$
$\displaystyle =A^{0}R^{0}=1$
23. If the first and the $\displaystyle n^{th}$ term of a G.P. are a and b respectively and if P is the product of n terms, prove that $\displaystyle P^{2}=(ab)^{n}$
Given,
first term is a, last term is b.
G.P. is $\displaystyle a,ar,ar^{2},...ar^{n-1}$
product $\displaystyle P=(a)(ar)(ar^{2})...(ar^{n-1})$
$\displaystyle =(a\times a\times ...a)(r\times r^{2}\times r^{3}...r^{n-1})$
$\displaystyle =(a^{n})(r^{1+2+...(n-1)})$
$\displaystyle =(a^{n})(r^{\frac{n(n-1)}{2}})$
$\displaystyle P^{2}=(a^{2n})(r^{n(n-1)})$
$\displaystyle P^{2}=[a^{2}r^{n-1}]^{n}$
$\displaystyle P^{2}=[a.ar^{n-1}]^{n}$
$\displaystyle P^{2}=[a.b]^{n}$
24. Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from $\displaystyle (n+1)^{th}$ to $\displaystyle (2n)^{th}$ term is $\displaystyle \frac{1}{r^{n}}$.
Given,
$\displaystyle S_{n}=\frac{a(1-r^{n})}{1-r}$ ...(i)
Since there are n terms from (n+1) to 2n,
$\displaystyle S_{n}=\frac{a_{n+1}(1-r^{n})}{1-r}$ $\displaystyle \Rightarrow S_{n}=\frac{ar^{n}(1-r^{n})}{1-r}$ ...(ii)
Required ratio is,
$\displaystyle \frac{(i)}{(ii)}$ $\displaystyle \Rightarrow \frac{\frac{a(1-r^{n})}{1-r}}{\frac{ar^{n}(1-r^{n})}{1-r}}$
$\displaystyle \frac{a(1-r^{n})}{1-r}\times \frac{1-r}{ar^{n}(1-r^{n})}$
$\displaystyle =\frac{1}{r^{n}}$
26. Intsert two numbers between 3 and 81 so that the resulting sequence is GP.
Let $\displaystyle a=3$
$\displaystyle a_{1},a_{2}$ are two numbers between 3 and 81
$\displaystyle a_{4}=81$ $\displaystyle \Rightarrow ar^{3}=81$
$\displaystyle 3r^{3}=81$
$\displaystyle r^{3}=27$ $\displaystyle \Rightarrow 3^{3}=27$
$\displaystyle r=3$
Therefore,
$\displaystyle a_{2}=ar=3\times 3=9$
$\displaystyle a_{3}=ar^{2}=3\times 3^{2}=27$
Two numbers between 3 and 81 are 9 and 27.
27. Find the value of n so that $\displaystyle \frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}$ may be the geometric mean between a and b.
We know that geometric mean between a and b is given by $\displaystyle \sqrt{ab}$
$\displaystyle \frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}=\sqrt{ab}$
squaring on both sides,
$\displaystyle \left (\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}} \right )^{2}=ab$
$\displaystyle \frac{a^{2(n+1)}+b^{2(n+1)}+2a^{n+1}b^{n+1}}{a^{2n}+b^{2n}+2a^{n}b^{n}}=ab$
$\displaystyle a^{2n+2}+b^{2n+2}+2a^{n+1}b^{n+1}$ $\displaystyle =ab(a^{2n}+b^{2n}+2a^{n}b^{n})$
$\displaystyle a^{2n+2}+b^{2n+2}+2a^{n+1}b^{n+1}$ $\displaystyle =a^{2n+1}b+ab^{2n+1}+2a^{n+1}b^{n+1}$
$\displaystyle a^{2n+2}+b^{2n+2}=a^{2n+1}b+ab^{2n+1}$
$\displaystyle a^{2n+2}-a^{2n+1}b=ab^{2n+1}-b^{2n+2}$
$\displaystyle a^{2n+1}(a-b)=b^{2n+1}(a-b)$
$\displaystyle a^{2n+1}=b^{2n+1}$
$\displaystyle \left ( \frac{a}{b} \right )^{2n+1}=1=\left ( \frac{a}{b} \right )^{0}$
$\displaystyle \Rightarrow 2n+1=0$
$\displaystyle n=-\frac{1}{2}$
28. The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio $\displaystyle (3+2\sqrt{2}):(3-2\sqrt{2})$.
Given,
$\displaystyle a+b=6\sqrt{ab}$ ...(i)
squaring on both sides,
$\displaystyle (a+b)^{2}=36ab$
we know that,
$\displaystyle (a-b)^{2}=(a+b)^{2}-4ab$
$\displaystyle (a-b)^{2}=36ab-4ab$
$\displaystyle (a-b)^{2}=32ab$
$\displaystyle a-b=\sqrt{32}\sqrt{ab}$
$\displaystyle a-b=4\sqrt{2}\sqrt{ab}$ ...(ii)
adding (i) and (ii)
$\displaystyle a+b+a-b=6\sqrt{ab}+4\sqrt{2}\sqrt{ab}$
$\displaystyle 2a=(6+4\sqrt{2})\sqrt{ab}$
$\displaystyle 2a=2(3+2\sqrt{2})\sqrt{ab}$
$\displaystyle a=(3+2\sqrt{2})\sqrt{ab}$
$\displaystyle (i)\Rightarrow b=6\sqrt{ab}-a$
$\displaystyle b=6\sqrt{ab}-(3+2\sqrt{2})\sqrt{ab}$
$\displaystyle b=(6-3-2\sqrt{2})\sqrt{ab}$
$\displaystyle b=(3-2\sqrt{2})\sqrt{ab}$
$\displaystyle \frac{a}{b}=\frac{(3+2\sqrt{2})\sqrt{ab}}{(3-2\sqrt{2})\sqrt{ab}}$ $\displaystyle =\frac{3+2\sqrt{2}}{3-2\sqrt{2}}$
Thus, $\displaystyle a:b=(3+2\sqrt{2}):(3-2\sqrt{2})$
29. If A and G be AM and GM respectively between two positive numbers, prove that the numbers are $\displaystyle A\pm \sqrt{(A+G)(A-G)}$
Let the two numbers be a and b
given that,
$\displaystyle AM=A=\frac{a+b}{2}$ $\displaystyle \Rightarrow a+b=2A$ ...(i)
$\displaystyle GM=G=\sqrt{ab}$ $\displaystyle \Rightarrow ab=G^{2}$
We know that,
$\displaystyle (a-b)^{2}=(a+b)^{2}-4ab$
$\displaystyle (a-b)^{2}=(2A)^{2}-4G^{2}$
$\displaystyle (a-b)^{2}=4A^{2}-4G^{2}$
$\displaystyle (a-b)^{2}=4(A^{2}-G^{2})$
$\displaystyle a-b=\pm \sqrt{4(A^{2}-G^{2})}$
$\displaystyle a-b=\pm 2\sqrt{(A-G)(A+G)}$
$\displaystyle a-(2A-a)$ $\displaystyle =\pm 2\sqrt{(A-G)(A+G)}$ [$\displaystyle b=2A-a$ ]
$\displaystyle a-2A+a=\pm 2\sqrt{(A-G)(A+G)}$
$\displaystyle 2a-2A=\pm 2\sqrt{(A-G)(A+G)}$
$\displaystyle 2a=2A\pm 2\sqrt{(A-G)(A+G)}$
$\displaystyle a=A\pm \sqrt{(A-G)(A+G)}$
$\displaystyle b=2A-A\pm \sqrt{(A-G)(A+G)}$
$\displaystyle b=A\pm \sqrt{(A-G)(A+G)}$
30. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of $\displaystyle 2^{nd}$ hour, $\displaystyle 4^{th}$ hour and $\displaystyle n^{th}$ hour?
Given that initial number of bacteria is 30 and doubles every hour.
∴ a=30, r=2
Number of bacteria at the end of $\displaystyle 2^{nd}$ hour, $\displaystyle a_{3}=ar^{2}=30(2)^{2}=120$
Number of bacteria at the end of $\displaystyle 4^{th}$ hour, $\displaystyle a_{5}=ar^{4}=30(2)^{4}=480$
Number of bacteria at the end of $\displaystyle n^{th}$ hour, $\displaystyle a_{n+1}=ar^{n}=30(2)^{n}$
31. What will Rs 500 amounts to 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?
Given that,
P = 500
r = 10
We know that, amount for compound interest, $\displaystyle A=P\left ( 1+\frac{r}{100} \right )^{n}$
amount at the end of first year = $\displaystyle 500\left ( 1+\frac{10}{100} \right )^{1}$ = $\displaystyle 500\left ( 1+\frac{1}{10} \right )$ = $\displaystyle 500\left ( \frac{11}{10} \right )$ = $\displaystyle 500(1.1)$
amount at the end of second year = $\displaystyle 500(1.1)(1.1)=500(1.1)^{2}$
amount at the end of third year = $\displaystyle 500(1.1)(1.1)(1.1)=500(1.1)^{3}$ so on..
amount at the end of 10 year = $\displaystyle 500(1.1)^{10}$
32. If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.
Let the roots of quadratic equation be a and b
Given that,
$\displaystyle AM=\frac{a+b}{2}=8 \Rightarrow a+b=16$ ...(i)
$\displaystyle GM=\sqrt{ab}=5 \Rightarrow ab=25$ ...(ii)
The quadratic equation is given by,
$\displaystyle x^{2}+(a+b)x+ab=0$
$\displaystyle x^{2}+16x+25=0$
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