11th NCERT/CBSE Sequences and series Exercise 9.3 Questions 32
Hi

Question (1)

Find the 20th and nth 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 12th term of G.P. whose 8th term is 192 and the common ratio is 2

Solution

Common ratio, r = 2 Let a be the first term of the G.P. ∴ a8 = ar8–1 = ar7
⇒ ar7 = 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 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps

Solution

a5 = p, a8 = 9, a11 = s
ar4 = p, ar7 = q, ar10 = s
RHS = ps
= ar4 . ar10 = a2r14
RHS = (ar7)2= q2 = LHS

Question (4)

The 4th term of G.P. is square of its second term, and the first term is -3 dtermine its 7th term

Solution

Let a be the first term and r be the common ratio of the G.P.
∴ a = –3
It is known that, an = arn–1
∴a4 = ar3 = (–3)r3
a2 = ar1 = (–3) r
According to the given condition,
(–3) r3 = [(–3) r]2
⇒ –3r3 = 9 r2
⇒ r = –3
a7 = a r7–1 = ar6 = (–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 nth 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, a2, -a3, ... (if a ≠ -1)

Solution

The given G.P. is 1, -a, a2, -a3, ...
Here, first term = a1 = 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 x3, x5, x7, .... ( if x ≠ ±1)

Solution

The given G.P. is x3, x5, x7, ....
Here, a = x3 and r = x2
\[{S_n} = \frac{{a\left( {1 - {r^n}} \right)}}{{1 - r}} = \frac{{{x^3}\left[ {1 - {{\left( {{x^2}} \right)}^n}} \right]}}{{1 - {r^2}}}\] \[{S_n} = \frac{{{x^3}\left( {1 - {x^{2n}}} \right)}}{{1 - {x^2}}}\]

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, 32, 33, … 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
a3 = 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, 32, 33, … are needed to give the sum 120?

Solution

The given G.P. is 3, 32, 33, …
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, ar2, ar3,...
According to the given condition,
a + ar + ar2 = 16 and ar3 + ar4 + ar5 = 128
⇒ a(1+r+r2) = 16 - - - (1)
ar3 (1+r+r2) = 128 - - - (2)
Dividing equation (2) by (1), we obtain $\frac{{a{r^3}\left( {1 + r + {r^2}} \right)}}{{a\left( {1 + r + {r^2}} \right)}} = \frac{{128}}{{16}}$
⇒ r3 = 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 7th term 64, determine S7.

Solution

a = 729
a7 = 64
Let r be the common ratio of the G.P.
It is known that, an = a rn–1
a7 = ar7–1 = (729)r6
⇒ 64 = 729 r6
\[ \Rightarrow {r^6} = \frac{{64}}{{729}}\] \[ \Rightarrow {r^6} = {\left( {\frac{2}{3}} \right)^6}\] \[ \Rightarrow r = \frac{2}{3}\] 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,
\[{S_4} = - 4 = \frac{{a\left( {1 - {r^2}} \right)}}{{1 - r}} - - - (1)\] a5 = 4 × a3
ar4 = 4ar2
⇒ r2 = 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 4th, 10th and 16th 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,
a4 = a r3 = x … (1)
a10 = a r9 = y … (2)
a16 = a r15 = 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 Sn = 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, ar2, ...,arn-1 and A, AR, AR2, ...,ARn-1 form a G.P, and find the common ratio.

Solution

It has to be proved that the sequence, aA, arAR, ar2AR2, …arn–1ARn–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 4th by 18.

Solution

Let 1st term of G.P. be 'a' and rate be 'r'
From given condition
a3 = a1 + 9
ar2 = a + 9
ar2 -a = 9
a(r2 - 1) = 9 ---(1)

a2 = a4 + 18
ar = ar3 + 18
ar - ar3 = 18
ar(1-r2) = 18
-ar(r2-1) = 18
ar(r2-1) = -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 pth, qth and rth terms of a G.P. are a, b and c, respectively. Prove that aq-r br-pcp-q = 1

Solution

Let A be the first term and R be the common ratio of the G.P.
According to the given information,
ARp–1 = a
ARq–1 = b
ARr–1 = c
aq–r br–p cp–q
= Aq–r × R(p–1) (q–r) × Ar–p × R(q–1) (r–p) × Ap–q × R(r –1)(p–q)
= Aq-r+r-p+p-q × R (pq – pr – q + r) + (rq – r + p – pq) + (pr – p – qr + q)
= A0 × R0
= 1
Thus, the given result is proved.

Question (23)

If the first and the nth term of a G.P. are a ad b, respectively, and if P is the product of n terms, prove that P2 = (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, ar2, ar3, … arn–1, where r is the common ratio.
b = arn–1 … (1)
P = Product of n terms
= (a) (ar) (ar2) … (arn–1)
= (a × a × .....a) (r × r2 ×....rn–1)
= an 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)}}$
an+1 = arn+1-1 = arn
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
(a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + dc)2

Solution

a, b, c, d are in G.P.
Let ratio be 'r'
a = a, b = ar, c = ar2, d = ar3
LHS=(a2 + b2 + c2) b2 + c2 + d2)
=[a2 + a2r2 + a2r4] [ a2r2 + a2r4+a2r6]
=a2(1+r2+r4)a2r2(1+r2+r4)
=a4r2(1+r2+r4)2
RHS = (ab + bc + cd)2
= [a(ar)+ (ar)(ar2) + (ar2)(ar3)]2
=[a2r + a2r3 + a2r5]2 =a4r2(1+r2+r4)2 = LHS

Question (26)

Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

Solution

Let G1 and G2 be two numbers between 3 and 81 such that the series, 3, G1, G2, 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
⇒ r3 = 27
∴ r = 3 (Taking real roots only)
For r = 3,
G1 = ar = (3) (3) = 9
G2 = ar2 = (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
$\frac{{{{\left( {{a^{n + 1}} + {b^{n + 1}}} \right)}^2}}}{{{{\left( {{a^n} + {b^n}} \right)}^2}}} = ab$
⇒ a2n+2 + 2an+1bn+1 + b2n+2 = (ab) (a2n + 2anbn+ b2n)
⇒ a2n+2 + 2an+1bn+1 + b2n+2 = a2n+1b + 2an+1 bn+1 + ab2n+1
⇒ a2n+2 +b2n+2 = a2n+1 b +ab2n+1
⇒ a2n+2 -a2n+2 b = ab2n+1 - b2n+2
⇒ a2n+1(a - b) = b2n+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 = G2
x(2A - x) = G2
2Ax - x2 = G2
x2 - 2Ax + G2 = 0
Comparing to std. quadric equation
a = 1, b = -A, c = G2
D = b2 -4ac
D= 4A2 - 4(1) (G2)
D= 4(A2 - G2)
$\sqrt D = 2\left( {\sqrt {{A^2} - {G^2}} } \right)$
$x = \frac{{ - b \pm \sqrt D }}{{2a}} = \frac{{2A \pm 2\sqrt {{A^2} - {G^2}} }}{{2\left( 1 \right)}}$
$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 2nd hour, 4th hour and nth 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
∴ a3 = ar2 = (30) (2)2 = 120
Therefore, the number of bacteria at the end of 2nd hour will be 120.
a5 = ar4 = (30) (2)4 = 480
The number of bacteria at the end of 4th hour will be 480.
an +1 = arn = (30) 2n
Thus, number of bacteria at the end of nth 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
an = arn-1
at end of 10, n = 11
a11 = a(1.1)11-1
a11 = 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,
x2– x (Sum of roots) + (Product of roots) = 0
x2 – x (a + b) + (ab) = 0
x2 – 16x + 25 = 0 [Using (1) and (2)]
Thus, the required quadratic equation is x2 – 16x + 25 = 0
Exercise 9.2⇐
⇒Exercise 9.4