An algebra problem by Rahil Sehgal

Algebra Level pending

The first term of a geometric progression is 1. The sum of its third and fifth terms is 90. Then find its common ratio.

± 2 \pm 2 ± 3 \pm 3 ± 5 \pm 5 ± 4 \pm 4

Rahil Sehgal by Rahil Sehgal , 20, India

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1 solution

Viki Zeta
Mar 18, 2017

a = 1 , r = ? , a n = a r n − 1 a 3 + a 5 = 90 r 2 + r 4 = 90 r 4 + r 2 − 90 = 0 ( r 2 + 10 ) ( r 2 − 9 ) = 0 r 2 = − 10 , r 2 = 9 r 2 = 9 r = ± 3 a = 1, r = ?, \\ a_n = ar^{n-1} \\ a_3 + a_5 = 90 \\ r^2 +r^4 = 90 \\ r^4 + r^2 - 90 = 0 \\ (r^2 + 10)(r^2 - 9) = 0 \\ r^2 = -10, ~ r^2= 9 \\ r^2 = 9 \\ \boxed{r = \pm 3}

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