Question

In a Geometric Progession the sum of 2nd and 4th term is 30.The difference of 6th and 2nd term is 90.Find the 8th term of a Geometric Progression whose common ratio is greater than 1.

Answer 1

ar+ar^3=30.
ar=30/(1+r^2)

ar^5-ar=90
ar(r^4-1)=90
30/(1+r^2)×(r^4-1)

now I think u can find r hence a and then the 8th term of the GP

Answer 2

ANSWER:-

$\sf{ ar + ar^3 = 30}\\\\\sf{ ar (1 + ar^2) }\\\\\sf{ar^2 + 1 = \dfrac{30}{ar} }\\\\\sf{ar^5 - ar = 90}\\\\\sf{ar(ar^4 - 1) = 90}\\\\\sf{ar^4 - 1 = \dfrac{90}{ar} }\\\\\sf{(ar^2 + 1)(ar^2 - 1) = 90}\\\\\sf{(\dfrac{30}{ar})(ar^2 - 1) = \dfrac{90}{ar}}\\\\\sf{r^2 - 1 = 3}\\\\\sf{r^2 = 4 \: and \: r = 2 }\\\\\sf{ar + ar^3 = 30}\\\\\sf{Therefore}\\\\\sf{T_8 = ar^7 = 384}$