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The sum of an infinite geometric series is $3$. A series, which is formed by squares of its terms, have the sum also $3$. First series will be
$\frac{3}{2},\frac{3}{4},\frac{3}{8},\frac{3}{{16}},.....$
$\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{{16}},.....$
$\frac{1}{3},\frac{1}{9},\frac{1}{{27}},\frac{1}{{81}},.....$
$1, - \frac{1}{3},\,\frac{1}{{{3^2}}}, - \frac{1}{{{3^3}}},.....$
Solution
(a) ${({S_1})_\infty } = \frac{a}{{1 – r}} = 3$ or $a = 3\,(\,1 – r)$ …..$(i)$
${({S_2})_\infty } = \frac{{{a^2}}}{{1 – {r^2}}} = 3$
or ${a^2} = 3\,(1 – {r^2})$ or $9\,{(1 – r)^2} = 3\,(1 – {r^2})$ [by $(i)$]
or $3\,(1 – 2r + {r^2}) = 1 – {r^2}$ or $2{r^2} – 3r + 1 = 0$
or $(r – 1)\,(2r – 1) = 0$,
$\therefore $$r = 1,\frac{1}{2}$
If $r = 1,$ then $a = 3(1 – 1) = 0$ which is impossible.
If $r = \frac{1}{2},$then $a = 3\,\left( {1 – \frac{1}{2}} \right) = 3/2$
So first series is $3/2, 3/4, 3/8, 3/16,…..$
