11th NCERT/CBSE Sequences and series Exercise 9.4(supplementary) Questions 6
Hi
Find the sum to infinity in each of the following G.P.

Question (1)

$1,\frac{1}{3},\frac{1}{9},.......$

Solution

It is G.P. with a = 1, $r = \frac{{\frac{1}{3}}}{1} = \frac{1}{3} < 1$
${S_\infty } = \frac{a}{{1 - r}}$
${S_\infty } = \frac{1}{{1 - \frac{1}{3}}} = \frac{1}{{\frac{2}{3}}} = \frac{3}{2} = 1.5$

Question (2)

6, 1.2, 0.24, .....

Solution

It is G.P. with a = 6, $r = \frac{{1.2}}{6} = \frac{{12}}{{60}} = 0.2$
${S_\infty } = \frac{a}{{1 - r}} = \frac{6}{{1 - 0.2}}$
${S_\infty } = \frac{6}{{0.8}} = \frac{{60}}{8} = 7.5$

Question (3)

$5,\frac{{20}}{7},\frac{{80}}{{49}},...$

Solution

It is G.P. with a=5, $r = \frac{{20}}{{7 \times 5}} = \frac{4}{7}$
${S_\infty } = \frac{a}{{1 - r}} = \frac{5}{{1 - \frac{4}{7}}}$
${S_\infty } = \frac{5}{{\frac{3}{7}}} = \frac{{35}}{3}$

Question (4)

$\frac{{ - 3}}{4},\frac{3}{{16}},\frac{{ - 3}}{{64}},....$

Solution

It is G.P. with $a = \frac{{ - 3}}{4},r = \frac{{\frac{3}{{16}}}}{{\frac{{ - 3}}{4}}} = - \frac{1}{4}$
${S_\infty } = \frac{a}{{1 - r}} = \frac{{\frac{{ - 3}}{4}}}{{1 + \frac{1}{4}}}$
${S_\infty } = \frac{{\frac{{ - 3}}{4}}}{{\frac{5}{4}}} = \frac{{ - 3}}{5}$

Question (5)

Prove that ${3^{\frac{1}{2}}} \times {3^{\frac{1}{4}}} \times {3^{\frac{1}{8}}} \times ...... = 3$

Solution

$LHS = {3^{\frac{1}{2}}} \times {3^{\frac{1}{4}}} \times {3^{\frac{1}{8}}} \times ......$
$LHS = {3^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + .....}}$
$\frac{1}{2}.\frac{1}{4},\frac{1}{8},....$ is G.P. with $a = \frac{1}{2},r = \frac{1}{2}$
${S_\infty } = \frac{a}{{1 - r}} = \frac{{\frac{1}{2}}}{{1 - \frac{1}{2}}} = \frac{{\frac{1}{2}}}{{\frac{1}{2}}} = 1$
∴ LHS = 31 = 3 = RHS

Question (6)

x = 1 + a + a2 + .... and y = 1 + b + b2+ ...., where |a| < 1 and |b| < 1 prove that
$1 + ab + {a^2}{b^2} = \frac{{xy}}{{x + y - 1}}$

Solution

x = 1 + a + a2 + ...
It is G.P. with a = 1, r = a
$x = \frac{a}{{1 - r}} = \frac{1}{{1 - a}}$
x(1-a) = 1
x - ax = 1
x - 1 = ax
$\frac{{x - 1}}{x} = a$
y = 1 + b + b2+ ....,
It is G.P. with a=1, r = b
${S_\infty } = \frac{a}{{1 - r}}$
$y = \frac{1}{{1 - b}}$
(1-b) y = 1
y-by = 1
y-1=by
$\frac{{y - 1}}{y} = b$

LHS = 1 + ab + a2b2 + ....
It is a = 1, r = ab
${S_\infty } = \frac{a}{{1 - r}} = \frac{1}{{1 - ab}}$
${S_\infty } = \frac{1}{{1 - \left( {\frac{{x - 1}}{x}} \right)\left( {\frac{{y - 1}}{y}} \right)}}$
${S_\infty } = \frac{{xy}}{{xy - \left( {x - 1} \right)\left( {y - 1} \right)}}$
${S_\infty } = \frac{{xy}}{{xy - xy + x + y - 1}}$
${S_\infty } = \frac{{xy}}{{x + y - 1}} = RHS$
Exercise9.4a⇐
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