Question

The ratio of sum of first three terms of a GP to the sum of first six terms of the GP is 64:91, the common ratio is
a.$\dfrac{1}{4}$
b.$\dfrac{3}{4}$
c.$\dfrac{5}{4}$
d.$\dfrac{7}{4}$

Answer 1

The sum of nth term of a GP is given by ${S_n} = \dfrac{{a({r^n} - 1)}}{{r - 1}}$ where a is the first term and r is the common ratio. In the question we are given a ratio of sum of first 3 terms to the sum of first 6 terms. We simply have to find the sums and equate it to ratio.

Complete step-by-step answer:
Let $a$ be the first term and $r$ be the common ratio of the given GP.
It is given that the ratio of sum of first 3 terms to the ratio of first 6 terms of the GP is 64:91
As we know that the Sum of n terms of a GP is given by
${S_n} = \dfrac{{a({r^n} - 1)}}{{r - 1}}$
$\Rightarrow \dfrac{{{S_3}}}{{{S_6}}} = \dfrac{{\dfrac{{a({r^3} - 1)}}{{r - 1}}}}{{\dfrac{{a({r^6} - 1)}}{{r - 1}}}}$
$\Rightarrow \dfrac{{64}}{{91}} = \dfrac{{{r^3} - 1}}{{{r^6} - 1}}$
$\Rightarrow 64({r^6} - 1) = 91({r^3} - 1)$
$\Rightarrow 64{r^6} - 64 = 91{r^3} - 91$
$\Rightarrow 64{r^6} - 91{r^3} + 27 = 0$
Let ${r^3} = t$
$\Rightarrow 64{t^2} - 91t + 27 = 0$
On solving this quadratic equation we get,
$t = \dfrac{{ - ( - 91) \pm \sqrt {{{( - 91)}^2} - 4(64)(27)} }}{{2(64)}}$
$\Rightarrow t = \dfrac{{91 \pm 37}}{{128}}$
$\Rightarrow t = \dfrac{{128}}{{128}}\& t = \dfrac{{54}}{{128}}$
$\Rightarrow t = 1\& t = \dfrac{{27}}{{64}}$
Substituting ${r^3} = t$
$\Rightarrow {r^3} = 1\& {r^3} = \dfrac{{27}}{{64}}$
$\Rightarrow r = 1\& r = \dfrac{3}{4}$
Hence the common ratio of the given GP is $r = \dfrac{3}{4}$.

Note: For any GP with common ratio r<1, the formula for the sum of nth term is given by,
${S_n} = \dfrac{{a(1 - {r^n})}}{{1 - r}}$
Also, the sum of an infinite GP is given by
${S_\infty } = \dfrac{a}{{1 - r}}$