Question

Geometric progression 10th term is 8 times the 13th term if the first term is 3 find the sum of infinite terms

Answer 1

Let first term of GP be a and common ratio be r.
Given a = 3
$nth \: term \: of \: gp \: is \: \: tn \: = \: a {r}^{n - 1} \\ given \: t10 = 8 \times t13 \\ i.e. \: a {r}^{9} = 8 \times a {r}^{12 } \\ {r}^{3} = \frac{1}{8} \: \: \: \: \: thus \: r = \frac{1}{2}$
$sum \: of \: infinite \: terms = \frac{a}{1 - r} \\ = \frac{3}{1 - \frac{1}{2} } = 6$