Question (1)
Find the 20
th and n
th terms of the G.P. $\frac{5}{2},\frac{5}{4},\frac{5}{8},....$
Solution
The given G.P. is $\frac{5}{2},\frac{5}{4},\frac{5}{8},....$
Here, a = First term =$\frac{5}{2}$
r = Common ratio = $\frac{{\frac{5}{4}}}{{\frac{5}{2}}} = \frac{1}{2}$
\[{a_{20}} = a{r^{20 - 1}} = \frac{5}{2}{\left( {\frac{1}{2}} \right)^{19}}\]
\[{a_{20}} = \frac{5}{{\left( 2 \right){{\left( 2 \right)}^{19}}}} = \frac{5}{{{{\left( 2 \right)}^{20}}}}\]
\[{a_n} = a{r^{n - 1}} = \frac{5}{2}{\left( {\frac{1}{2}} \right)^{n - 1}}\]
\[{a_n} = \frac{5}{{\left( 2 \right){{\left( 2 \right)}^{n - 1}}}} = \frac{5}{{{{\left( 2 \right)}^n}}}\]
Question (2)
Find the 12
th term of G.P. whose 8
th term is 192 and the common ratio is 2
Solution
Common ratio, r = 2
Let a be the first term of the G.P.
∴ a
8 = ar
8–1 = ar
7
⇒ ar
7 = 192
a(2)
7 = 192
a(2)
7 = (2)
6 (3)
$ \Rightarrow a = \frac{{{{\left( 2 \right)}^6} \times 3}}{{{{\left( 2 \right)}^7}}} = \frac{3}{2}$
\[ \therefore {a_{12}} = a{r^{12 - 1}} = \left( {\frac{3}{2}} \right){\left( 2 \right)^{11}} = \left( 3 \right){\left( 2 \right)^{10}} = 3072\]
Question (3)
The 5
th, 8
th and 11
th terms of a G.P. are
p,
q and
s, respectively. Show that
q2 =
ps
Solution
a
5 = p, a
8 = 9, a
11 = s
ar
4 = p, ar
7 = q, ar
10 = s
RHS = ps
= ar
4 . ar
10 = a
2r
14
RHS = (ar
7)
2= q
2 = LHS
Question (4)
The 4
th term of G.P. is square of its second term, and the first term is -3 dtermine its 7
th term
Solution
Let a be the first term and r be the common ratio of the G.P.
∴ a = –3
It is known that, a
n = ar
n–1
∴a
4 = ar
3 = (–3)r
3
a
2 = ar
1 = (–3) r
According to the given condition,
(–3) r
3 = [(–3) r]
2
⇒ –3r
3 = 9 r
2
⇒ r = –3
a
7 = a r
7–1 = ar
6 = (–3)(–3)
6 = – (3)
7 = –2187
Thus, the seventh term of the G.P. is –2187.
Question (5)
(a) Which term of the following sequences: 2, 2√2, 4, ... is 128?
Solution
(a) The given sequence is 2, 2√2, 4, ...
Here a = 2 and $r = \frac{{2\sqrt 2 }}{2} = \sqrt 2 $
Let the nth term of the given sequence be 128.
\[{a_n} = a{r^{n - 1}}\]
\[ \Rightarrow \left( 2 \right){\left( {\sqrt 2 } \right)^{n - 1}} = 128\]
\[ \Rightarrow \left( 2 \right){\left( 2 \right)^{\frac{{n - 1}}{2}}} = {\left( 2 \right)^7}\]
\[ \Rightarrow {\left( 2 \right)^{\frac{{n - 1}}{2} + 1}} = {\left( 2 \right)^7}\]
\[ \therefore \frac{{n - 1}}{2} + 1 = 7\]
\[ \Rightarrow \frac{{n - 1}}{2} = 6\]
\[ \Rightarrow n - 1 = 12\]
\[ \Rightarrow n = 13\]
Thus, the 13th term of the given sequence is 128.
(b) Which term of the following sequences: √3, 3, 3√3, .... is 729?
Solution
The given sequence is √3, 3, 3√3, ...
Here, a= √3 and $r = \frac{3}{{\sqrt 3 }} = \sqrt 3 $
Let the n
th term of the given sequence be 729.
\[{a_n} = a{r^{n - 1}}\]
\[a{r^{n - 1}} = 729\]
\[ \Rightarrow \left( {\sqrt 3 } \right){\left( {\sqrt 3 } \right)^{n - 1}} = 729\]
\[ \Rightarrow {\left( 3 \right)^{\frac{1}{2}}}{\left( 3 \right)^{\frac{{n - 1}}{2}}} = {\left( 3 \right)^6}\]
\[ \Rightarrow {\left( 3 \right)^{\frac{1}{2} + \frac{{n - 1}}{2}}} = {\left( 3 \right)^6}\]
\[\frac{1}{2} + \frac{{n - 1}}{2} = 6\]
\[ \Rightarrow \frac{{1 + n - 1}}{2} = 6\]
\[ \Rightarrow n = 12\]
Thus, the 12th term of the given sequence is 729.
(c) Which term of the following sequences: $\frac{1}{3},\frac{1}{9},\frac{1}{{27}}...\;is \; \frac{1}{{19683}}?$
Solution
The given sequence is $\frac{1}{3},\frac{1}{9},\frac{1}{{27}}...\;$
Here, $a = \frac{1}{3} \; and \;r = \frac{1}{9} \div \frac{1}{3} = \frac{1}{3}$
Let the nth term of the given sequence be $\frac{1}{{19683}}$
\[{a_n} = a{r^{n - 1}}\]
\[ \therefore a{r^{n - 1}} = \frac{1}{{19683}}\]
\[ \Rightarrow \left( {\frac{1}{3}} \right){\left( {\frac{1}{3}} \right)^{n - 1}} = \frac{1}{{19683}}\]
\[ \Rightarrow {\left( {\frac{1}{3}} \right)^n} = {\left( {\frac{1}{3}} \right)^9}\]
\[ \Rightarrow n = 9\]
Thus, the 9th term of the given sequence is $\frac{1}{{19683}}$
Question (6)
For what values of x, the numbers $\frac{2}{7},x,\frac{{ - 7}}{2}$ are in G.P.
Solution
The given numbers are $\frac{2}{7},x,\frac{{ - 7}}{2}$
Common ration $ = \frac{{\frac{x}{{ - 2}}}}{7} = \frac{{ - 7x}}{2}$
Also common ratio $ = \frac{{\frac{{ - 7}}{2}}}{x} = \frac{{ - 7}}{{2x}}$
\[ \therefore \frac{{ - 7x}}{2} = \frac{{ - 7}}{{2x}}\]
\[ \Rightarrow {x^2} = \frac{{ - 2 \times 7}}{{ - 2 \times 7}} = 1\]
\[ \Rightarrow x = \sqrt 1 \]
\[ \Rightarrow x = \pm 1\]
Thus, for x = ± 1, the given numbers will be in G.P.
Question (7)
Find the sum to 20 terms in the geometric progression 0.15, 0.015, 0.0015 …
Solution
The given G.P. is 0.15, 0.015, 0.00015,
Here, a = 0.15 and $r = \frac{{0.015}}{{0.15}} = 0.1$
\[{S_n} = \frac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}\]
\[{S_{20}} = \frac{{0.15\left[ {1 - {{\left( {0.1} \right)}^{20}}} \right]}}{{1 - 0.1}}\]
\[{S_{20}} = \frac{{0.15}}{{0.9}}\left[ {1 - {{\left( {0.1} \right)}^{20}}} \right]\]
\[{S_{20}} = \frac{{15}}{{90}}\left[ {1 - {{\left( {0.1} \right)}^{20}}} \right]\]
\[{S_{20}} = \frac{1}{6}\left[ {1 - {{\left( {0.1} \right)}^{20}}} \right]\]
Question (8)
Find the sum to n terms in the geometric progression $\sqrt 7 ,\sqrt {21} ,3\sqrt 7 ...$
Solution
The given G.P. is $\sqrt 7 ,\sqrt {21} ,3\sqrt 7 ...$
Here, $a = \sqrt 7 $ and $r = \frac{{\sqrt {21} }}{{\sqrt 7 }} = \sqrt 3 $
\[{S_n} = \frac{{a\left( { {r^n}-1} \right)}}{{r -1}}\]
\[{S_n} = \frac{{\sqrt 7 \left[ {{{\left( {\sqrt 3 } \right)}^n} - 1} \right]}}{{\sqrt 3 - 1}}\]
By rationalizing
\[{S_n} = \frac{{\sqrt 7 \left[ {{{\left( {\sqrt 3 } \right)}^n} - 1} \right]}}{{\sqrt 3 - 1}} \times \frac{{1 + \sqrt 3 }}{{1 + \sqrt 3 }}\]
\[{S_n} = \frac{{\sqrt 7 \left( {1 + \sqrt 3 } \right)\left[ {{{\left( {\sqrt 3 } \right)}^n} - 1} \right]}}{{3 - 1}}\]
\[{S_n} = \frac{{\sqrt 7 \left( {1 + \sqrt 3 } \right)}}{2}\left[ {{{\left( {\sqrt 3 } \right)}^n} - 1} \right]\]
Question (9)
Find the sum to n terms in the geometric progression 1, -a, a
2, -a
3, ... (if a ≠ -1)
Solution
The given G.P. is 1, -a, a
2, -a
3, ...
Here, first term = a
1 = 1
Common ratio = r = – a
\[{S_n} = \frac{{{a_1}\left( {1 - {r^n}} \right)}}{{1 - r}}\]
\[{S_n} = \frac{{{a_1}\left[ {1 - {{\left( { - a} \right)}^n}} \right]}}{{1 - r}}\]
\[\therefore {S_n} = \frac{{1\left[ {1 - {{\left( { - a} \right)}^n}} \right]}}{{1 - \left( { - a} \right)}} = \frac{{\left[ {1 - {{\left( { - a} \right)}^n}} \right]}}{{1 + a}}\]
Question (10)
Find the sum to n terms in the geometric progression x
3, x
5, x
7, .... ( if x ≠ ±1)
Solution
The given G.P. is x
3, x
5, x
7, ....
Here, a = x
3 and r = x
2
\[{S_n} = \frac{{a\left( {1 - {r^n}} \right)}}{{1 - r}} = \frac{{{x^3}\left[ {1 - {{\left( {{x^2}} \right)}^n}} \right]}}{{1 - {r^2}}}\]
\[{S_n} = \frac{{{x^3}\left( {1 - {x^{2n}}} \right)}}{{1 - {x^2}}}\]
Question (11)
Evaluate $\sum\limits_{k = 1}^{11} {\left( {2 + {3^k}} \right)} $
Solution
\[\sum\limits_{k = 1}^{11} {\left( {2 + {3^k}} \right)} = \sum\limits_{k = 1}^{11} {\left( 2 \right)} + \sum\limits_{k = 1}^{11} {\left( {{3^k}} \right)} = 2\left( {11} \right) + \sum\limits_{k = 1}^{11} {\left( {{3^k}} \right)} \]
\[\sum\limits_{k = 1}^{11} {\left( {2 + {3^k}} \right)} = 22 + \sum\limits_{k = 1}^{11} {\left( {{3^k}} \right)} - - - (1)\]
\[\sum\limits_{k = 1}^{11} {\left( {{3^k}} \right)} = {3^1} + {3^2} + {3^3} + ... + {3^n}\]
The terms of this sequence 3, 3
2, 3
3, … forms a G.P.
\[{S_n} = \frac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}\]
\[ \Rightarrow {S_n} = \frac{{3\left[ {{{\left( 3 \right)}^{11}} - 1} \right]}}{{3 - 1}}\]
\[ \Rightarrow {S_n} = \frac{3}{2}\left( {{3^{11}} - 1} \right)\]
\[\therefore \sum\limits_{k = 1}^{11} {{3^k}} = \frac{3}{2}\left( {{3^{11}} - 1} \right)\]
Substituting this value in (1), we obtain
\[\sum\limits_{k = 1}^{11} {\left( {2 + {3^k}} \right)} = 22 + \frac{3}{2}\left( {{3^{11}} - 1} \right)\]
Question (12)
The sum of first three terms of a G.P. $\frac{{39}}{{10}}$ is and their product is 1. Find the common ratio and the terms.
Solution
Let $\frac{a}{r}$, a, ar be the first trhree terms of the G.P.
\[\frac{a}{r} + a + ar = \frac{{39}}{{10}} - - - (1)\]
\[\left( {\frac{a}{r}} \right)\left( a \right)\left( {ar} \right) = 1 - - - (2)\]
From (2), we obtain
a
3 = 1
⇒ a = 1 (Considering real roots only)
Substituting a = 1 in equation (1), we obtain
\[\frac{1}{r} + 1 + r = \frac{{39}}{{10}}\]
\[ \Rightarrow 1 + r + {r^2} = \frac{{39}}{{10}}r\]
\[ \Rightarrow 10 + 10r + 10{r^2} = 39r\]
\[ \Rightarrow 10{r^2} - 29r + 10 = 0\]
\[ \Rightarrow 10{r^2} - 25r - 4r + 10 = 0\]
\[ \Rightarrow 5r\left( {2r - 5} \right) - 2\left( {2r - 5} \right) = 0\]
\[ \Rightarrow \left( {5r - 2} \right)\left( {2r - 5} \right) = 0\]
\[ \Rightarrow r = \frac{2}{5} \;or \;\frac{5}{2}\]
Thus, the three terms of G.P. are $\frac{5}{2},1,\;and\; \frac{2}{5}$
Question (13)
How many terms of G.P. 3, 3
2, 3
3, … are needed to give the sum 120?
Solution
The given G.P. is 3, 3
2, 3
3, …
Let n terms of this G.P. be required to obtain the sum as 120.
\[{S_n} = \frac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}\]
Here, a = 3 and r = 3
\[{S_n} = 120 = \frac{{3\left( {{3^n} - 1} \right)}}{{3 - 1}}\]
\[ \Rightarrow 120 = \frac{{3\left( {{3^n} - 1} \right)}}{2}\]
\[ \Rightarrow \frac{{120 \times 2}}{3} = {3^n} - 1\]
\[ \Rightarrow {3^n} - 1 = 80\]
\[ \Rightarrow {3^n} = 81\]
\[ \Rightarrow {3^n} = {3^4}\]
\[n = 4\]
Thus, four terms of the given G.P. are required to obtain the sum as 120
Question (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.
Solution
Let the G.P. be a, ar, ar
2, ar
3,...
According to the given condition,
a + ar + ar
2 = 16 and ar
3 + ar
4 + ar
5 = 128
⇒ a(1+r+r
2) = 16 - - - (1)
ar
3 (1+r+r
2) = 128 - - - (2)
Dividing equation (2) by (1), we obtain
$\frac{{a{r^3}\left( {1 + r + {r^2}} \right)}}{{a\left( {1 + r + {r^2}} \right)}} = \frac{{128}}{{16}}$
⇒ r
3 = 8
∴ r = 2
Substituting r = 2 in (1), we obtain
a (1 + 2 + 4) = 16
⇒ a (7) = 16
$ \Rightarrow a = \frac{{16}}{7}$
${S_n} = \frac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$
\[ \Rightarrow {S_n} = \frac{{16}}{7}\frac{{\left( {{2^n} - 1} \right)}}{{2 - 1}} = \frac{{16}}{7}\left( {{2^n} - 1} \right)\]
Question (15)
Given a G.P. with a = 729 and 7
th term 64, determine S7.
Solution
a = 729
a
7 = 64
Let r be the common ratio of the G.P.
It is known that, a
n = a r
n–1
a7 = ar
7–1 = (729)r
6
⇒ 64 = 729 r
6
\[ \Rightarrow {r^6} = \frac{{64}}{{729}}\]
\[ \Rightarrow {r^6} = {\left( {\frac{2}{3}} \right)^6}\]
\[ \Rightarrow r = \frac{2}{3}\]
Also, it is known that,
\[{S_n} = \frac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}\]
\[{S_7} = \frac{{729\left[ {1 - {{\left( {\frac{2}{3}} \right)}^7}} \right]}}{{1 - \frac{2}{3}}}\]
\[{S_7} = 3 \times 729\left[ {1 - {{\left( {\frac{2}{3}} \right)}^7}} \right]\]
\[{S_7} = {\left( 3 \right)^7}\left[ {\frac{{{{\left( 3 \right)}^7} - {{\left( 2 \right)}^7}}}{{{{\left( 3 \right)}^7}}}} \right]\]
\[{S_7} = {\left( 3 \right)^7} - {\left( 2 \right)^7}\]
\[{S_7} = 2187 - 128 = 2059\]
Question (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.
Solution
Let a be the first term and r be the common ratio of the G.P.
According to the given conditions,
\[{S_4} = - 4 = \frac{{a\left( {1 - {r^2}} \right)}}{{1 - r}} - - - (1)\]
a
5 = 4 × a
3
ar
4 = 4ar
2
⇒ r
2 = 4
∴ r = ± 2
From (1), we obtain
r = 2 > 1
\[{S_2} = \frac{{a\left( {{r^2} - 1} \right)}}{{r - 1}}\]
\[ - 4 = \frac{{a\left( {{2^2} - 1} \right)}}{{2 - 1}}\]
\[a = \frac{{ - 4}}{3}\]
\[ar = \frac{{ - 4}}{3} \times 2 = \frac{{ - 8}}{3}\]
\[GP\;is\;\frac{{ - 4}}{3},\frac{{ - 8}}{3},\frac{{ - 16}}{3},...\]
\[If\;r = - 2 < 1\]
\[{S_2} = \frac{{a\left( {1 - {r^2}} \right)}}{{1 - r}}\]
\[ - 4 = \frac{{a\left[ {1 - {{\left( { - 2} \right)}^2}} \right]}}{{\left[ {1 - \left( { - 2} \right)} \right]}}\]
\[ - 4 = \frac{{a\left( { - 3} \right)}}{3}\]
\[a = 4\]
\[ar = 4 \times \left( { - 2} \right) = - 8\]
\[GP\;is\;4, - 8,16,....\]
Question (17)
If the 4
th, 10
th and 16
th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.
Solution
Let a be the first term and r be the common ratio of the G.P.
According to the given condition,
a
4 = a r
3 = x … (1)
a
10 = a r
9 = y … (2)
a
16 = a r
15 = z … (3)
Dividing (2) by (1), we obtain
$\frac{y}{x} = \frac{{a{r^9}}}{{a{r^3}}} \Rightarrow \frac{y}{x} = {r^6}$
Dividing (3) by (2), we obtain
$\frac{z}{y} = \frac{{a{r^{15}}}}{{a{r^9}}} \Rightarrow \frac{z}{y} = {r^6}$
\[ \therefore \frac{y}{x} = \frac{z}{y}\]
Thus, x, y, z are in G. P.
Question (18)
Find the sum to n terms of the sequence, 8, 88, 888, 8888…
Solution
The given sequence is 8, 88, 888, 8888…
This sequence is not a G.P. However, it can be changed to G.P. by writing the terms as
S
n = 8 + 88 + 888 + 8888 + …………….. to n terms
\[{S_n} = \frac{8}{9}\left[ {9 + 99 + 999 + 9999 + ....\text{to n terms}} \right]\]
\[{S_n} = \frac{8}{9}\left[ {\left( {10 - 1} \right) + \left( {{{10}^2} - 1} \right) + \left( {{{10}^3} - 1} \right) + \left( {{{10}^4} - 1} \right) + ....\text{to n terms]} \right]\]
\[{S_n} = \frac{8}{9}\left[ {\left( {10 + {{10}^2} + ...\text{n terms}} \right) - \left( {1 + 1 + 1 + ...\text{n terms}} \right)} \right]\]
\[{S_n} = \frac{8}{9}\left[ {\frac{{10\left( {{{10}^n} - 1} \right)}}{{10 - 1}} - n} \right]\]
\[{S_n} = \frac{8}{9}\left[ {\frac{{10\left( {{{10}^n} - 1} \right)}}{9} - n} \right]\]
\[{S_n} = \frac{{80}}{{81}}\left( {{{10}^n} - 1} \right) - \frac{8}{9}n\]
Question (19)
Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, $\frac{1}{2}$
Solution
Required sum = $2 \times 128 + 4 \times 32 + 8 \times 8 + 16 \times 2 + 32 \times \frac{1}{2}$
$ = 256 + 128 + 64 + 32 + ...$
a = 256, and $r = \frac{{128}}{{256}} = \frac{1}{2} < 1$, n =5
${S_n} = \frac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}$
${S_n} = \frac{{256\left[ {1 - {{\left( {\frac{1}{2}} \right)}^5}} \right]}}{{1 - \left( {\frac{1}{2}} \right)}}$
${S_n} = \frac{{256}}{{\frac{1}{2}}}\left[ {1 - \frac{1}{{32}}} \right]$
${S_n} = \require{cancel} \cancel{256}^8 \times 2 \times \frac{{31}}{{\cancel{32}}}$
=496
Question (20)
Show that the products of the corresponding terms of the sequences a, ar, ar
2, ...,ar
n-1 and A, AR, AR
2, ...,AR
n-1 form a G.P, and find the common ratio.
Solution
It has to be proved that the sequence, aA, arAR, ar
2AR
2, …ar
n–1AR
n–1, forms a G.P.
$\frac{{ \text{Second Term}}}{{\text{First Term}}} = \frac{{arAR}}{{aA}} = rR$
$\frac{{\text{Third Term}}}{{\text{Second Term}}} = \frac{{a{r^2}A{R^2}}}{{arAR}} = rR$
Thus, the above sequence forms a G.P. and the common ratio is rR.
Question (21)
Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and the second term is greater than the 4
th by 18.
Solution
Let 1
st term of G.P. be 'a' and rate be 'r'
From given condition
a
3 = a
1 + 9
ar
2 = a + 9
ar
2 -a = 9
a(r
2 - 1) = 9 ---(1)
a
2 = a
4 + 18
ar = ar
3 + 18
ar - ar
3 = 18
ar(1-r
2) = 18
-ar(r
2-1) = 18
ar(r
2-1) = -18 ---(2)
(2) ÷ (1)
$\frac{{ar\left( {{r^2} - 1} \right)}}{{a\left( {{r^2} - 1} \right)}} = \frac{{ - 18}}{9}$
∴ r = -2
Substituting r = -2 in equation(1)
[(-2)
2 - 1 ] = 9
∴ 3a = 9
a = 3
So G.P. is 3, -6, 12, -24, .....
Question (22)
If the p
th, q
th and r
th terms of a G.P. are a, b and c, respectively. Prove that a
q-r b
r-pc
p-q = 1
Solution
Let A be the first term and R be the common ratio of the G.P.
According to the given information,
AR
p–1 = a
AR
q–1 = b
AR
r–1 = c
a
q–r b
r–p c
p–q
= A
q–r × R
(p–1) (q–r) × A
r–p × R
(q–1) (r–p) × A
p–q × R
(r –1)(p–q)
= A
q-r+r-p+p-q × R
(pq – pr – q + r) + (rq – r + p – pq) + (pr – p – qr + q)
= A
0 × R
0
= 1
Thus, the given result is proved.
Question (23)
If the first and the n
th term of a G.P. are a ad b, respectively, and if P is the product of n terms, prove that P
2 = (ab)
n.
Solution
The first term of the G.P is a and the last term is b.
Therefore, the G.P. is a, ar, ar
2, ar
3, … ar
n–1, where r is the common ratio.
b = ar
n–1 … (1)
P = Product of n terms
= (a) (ar) (ar
2) … (ar
n–1)
= (a × a × .....a) (r × r
2 ×....r
n–1)
= a
n r
1 + 2 +…(n–1) … (2)
Here, 1, 2, …(n – 1) is an A.P.
∴ 1 + 2 + .....+ (n-1)
$ = \frac{{n - 1}}{2}\left[ {2 + \left( {n - 1 - 1} \right) \times 1} \right]$
$ = \frac{{n - 1}}{2}\left[ {2 + n - 2} \right] = \frac{{n\left( {n - 1} \right)}}{2}$
$P = {a^n}{r^{\frac{{n\left( {n - 1} \right)}}{2}}}$
${P^2} = {a^{2n}}{r^{n\left( {n - 1} \right)}}$
${P^2} = {\left[ {{a^2}{r^{\left( {n - 1} \right)}}} \right]^n}$
${P^2} = {\left[ {a \times a \times {r^{\left( {n - 1} \right)}}} \right]^n}$
Using (1)
\[{P^2} = {\left( {ab} \right)^n}\]
Thus, the given result is proved.
Question (24)
Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from
${\left( {n + 1} \right)^{th}} \; to \;{\left( {2n} \right)^{th}}$ term is $\frac{1}{{{r^n}}}$
Solution
Let a be the first term and r be the common ratio of the G.P.
Sum of first n term $ = \frac{{a\left( {1 - {r^n}} \right)}}{{\left( {1 - r} \right)}}$
Since there are n terms from (n +1)
th to (2n)
th term,
Sum of terms from(n + 1)
th to (2n)
th term $ = \frac{{{a_{n + 1}}\left( {1 - {r^n}} \right)}}{{\left( {1 - r} \right)}}$
a
n+1 = ar
n+1-1 = ar
n
Thus, required ratio $ = \frac{{a\left( {1 - {r^n}} \right)}}{{\left( {1 - r} \right)}} \times \frac{{\left( {1 - r} \right)}}{{a{r^n}\left( {1 - {r^n}} \right)}} = \frac{1}{{{r^n}}}$
Thus, the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)
th to (2n)
th term is $\frac{1}{{{r^n}}}$
Question (25)
If a, b, c and d are in G.P. show that
(a
2 + b
2 + c
2) (b
2 + c
2 + d
2) = (ab + bc + dc)
2
Solution
a, b, c, d are in G.P.
Let ratio be 'r'
a = a, b = ar, c = ar
2, d = ar
3
LHS=(a
2 + b
2 + c
2) b
2 + c
2 + d
2)
=[a
2 + a
2r
2 + a
2r
4] [ a
2r
2 + a
2r
4+a
2r
6]
=a
2(1+r
2+r
4)a
2r
2(1+r
2+r
4)
=a
4r
2(1+r
2+r
4)
2
RHS = (ab + bc + cd)
2
= [a(ar)+ (ar)(ar
2) + (ar
2)(ar
3)]
2
=[a
2r + a
2r
3 + a
2r
5]
2
=a
4r
2(1+r
2+r
4)
2 = LHS
Question (26)
Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
Solution
Let G
1 and G
2 be two numbers between 3 and 81 such that the series, 3, G
1, G
2, 81, forms a G.P.
Let a be the first term and r be the common ratio of the G.P.
∴81 = (3) (r)
3
⇒ r
3 = 27
∴ r = 3 (Taking real roots only)
For r = 3,
G
1 = ar = (3) (3) = 9
G
2 = ar
2 = (3) (3)
2 = 27
Thus, the required two numbers are 9 and 27.
Question (27)
Find the value of n so that $\frac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}}$ may be the geometric mean between a and b.
Solution
G. M. of a and b is $\sqrt {ab} $
By the given condition $\frac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}} = \sqrt {ab} $
Squaring both sides, we obtain
$\frac{{{{\left( {{a^{n + 1}} + {b^{n + 1}}} \right)}^2}}}{{{{\left( {{a^n} + {b^n}} \right)}^2}}} = ab$
⇒ a
2n+2 + 2a
n+1b
n+1 + b
2n+2 = (ab) (a
2n + 2a
nb
n+ b
2n)
⇒ a
2n+2 + 2a
n+1b
n+1 + b
2n+2 =
a
2n+1b + 2a
n+1 b
n+1 + ab
2n+1
⇒ a
2n+2 +b
2n+2 = a
2n+1 b +ab
2n+1
⇒ a
2n+2 -a
2n+2 b = ab
2n+1 - b
2n+2
⇒ a
2n+1(a - b) = b
2n+1(a-b)
$ \Rightarrow {\left( {\frac{a}{b}} \right)^{2n + 1}} = 1 = {\left( {\frac{a}{b}} \right)^0}$
$ \Rightarrow 2n + 1 = 0$
$ \Rightarrow n = \frac{{ - 1}}{2}$
Question (28)
The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio $\left( {3 + 2\sqrt 2 } \right):\left( {3 - 2\sqrt 2 } \right)$
Solution
let numbers be x and y
$G.M. = \sqrt {xy} $
According to condition
$x + y = 6\sqrt {xy} $
$\frac{{x + y}}{{2\sqrt {xy} }} = \frac{3}{1}$
By comp. and divd. prop
$\frac{{x + y + 2\sqrt {xy} }}{{x + y - 2\sqrt {xy} }} = \frac{{3 + 1}}{{3 - 1}}$
$\frac{{{{\left( {\sqrt x + \sqrt y } \right)}^2}}}{{{{\left( {\sqrt x - \sqrt y } \right)}^2}}} = \frac{4}{2} = 2$
\[\frac{{\sqrt x + \sqrt y }}{{\sqrt x - \sqrt y }} = \frac{{\sqrt 2 }}{1}\]
By comp. and divd. prop
$\frac{{2\sqrt x }}{{2\sqrt y }} = \frac{{\sqrt 2 + 1}}{{\sqrt 2 - 1}}$
Squaring
$\frac{x}{y} = \frac{{{{\left( {\sqrt 2 + 1} \right)}^2}}}{{{{\left( {\sqrt 2 - 1} \right)}^2}}}$
$ = \frac{{2 + 1 + 2\sqrt 2 }}{{2 + 1 - 2\sqrt 2 }}$
$ = \frac{{3 + 2\sqrt 2 }}{{3 - 2\sqrt 2 }}$
Question (29)
If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are $A \pm \sqrt {\left( {A + G} \right)\left( {A - G} \right)} $
Solution
Let two number be x and y
then $A.M = \frac{{x + y}}{2} = A$
x+y = 2A
y = (2A-x) ... (1)
$GM = G = \sqrt {xy} $
∴ xy = G
2
x(2A - x) = G
2
2Ax - x
2 = G
2
x
2 - 2Ax + G
2 = 0
Comparing to std. quadric equation
a = 1, b = -A, c = G
2
D = b
2 -4ac
D= 4A
2 - 4(1) (G
2)
D= 4(A
2 - G
2)
$\sqrt D = 2\left( {\sqrt {{A^2} - {G^2}} } \right)$
$x = \frac{{ - b \pm \sqrt D }}{{2a}} = \frac{{2A \pm 2\sqrt {{A^2} - {G^2}} }}{{2\left( 1 \right)}}$
$x = A \pm \sqrt {{A^2} - {G^2}} $
$y = A \mp \sqrt {{A^2} - {G^2}} $
Question (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 2
nd hour, 4
th hour and n
th hour?
Solution
It is given that the number of bacteria doubles every hour. Therefore, the number of bacteria after every hour will form a G.P.
Here, a = 30 and r = 2
∴ a
3 = ar
2 = (30) (2)
2 = 120
Therefore, the number of bacteria at the end of 2nd hour will be 120.
a
5 = ar
4 = (30) (2)
4 = 480
The number of bacteria at the end of 4th hour will be 480.
a
n +1 = ar
n = (30) 2
n
Thus, number of bacteria at the end of n
th hour will be 30(2)
n.
Question (31)
What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?
Solution
a= 500
rate of interest R = 10%
$A = a{\left( {1 + \frac{R}{{100}}} \right)^n}$
When n = 1
A = 500(1.1)
When n = 2
A = 500(1.1)
2
Ratio r = 1.1
a
n = ar
n-1
at end of 10, n = 11
a
11 = a(1.1)
11-1
a
11 = 500(1.1)
10
Question (32)
If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.
Solution
Let the root of the quadratic equation be a and b.
According to the given condition,
$A.M. = \frac{{a + b}}{2} = 8 \Rightarrow a + b = 16 - - - (1)$
$G.M. = \sqrt {ab} = 5 \Rightarrow ab = 25 - - - (2)$
The quadratic equation is given by,
x
2– x (Sum of roots) + (Product of roots) = 0
x
2 – x (a + b) + (ab) = 0
x
2 – 16x + 25 = 0 [Using (1) and (2)]
Thus, the required quadratic equation is x
2 – 16x + 25 = 0