Question

In a geometric progression the sum of 2 and 4 term is 30. The difference of 6&2 term is 90. Find the 8 term of geometric progression, whose common ratio is greater then 1 .

Answer 1

Its easy...ar + ar3 = 30,  ar(1+ar2) = 30,  ar2 + 1=30/ar.

ar5-ar=90,  ar( ar4-1) = 90, ar4-1=90/ar , (ar2+1) (ar2-1) = 90, (30/ar)(ar2-1) = 90/ar, r2 - 1=3, r2 = 4 and r = 2....

a=3 by substituting r=2 in the equation ar + ar3= 30.. then T8=ar7=384. 

Answer 2

Given :
nth term of G.P=an=arⁿ⁻¹

second term=ar

4th term=ar³

6th term=ar⁵

Sum of 2 and 4th term =30

ar+ar³=20------------equation i

ar(1+r²)=30-----------equation iv

now difference of 6th and 2term is 90

ar⁵-ar=90------------------equation ii

Solving equation 1 and 2
 we get

ar+ar³=30
ar⁵-ar=90
+  +     +
________

ar³+ar⁵=120 ---------equation iii

ar³(1+r2)=120----------equation v

ar(1+r2)=30 ----------equation IV

by solving equation Iv and V( divide )

ar³/ar=120/30

r²=4

r=2

By substituting value of r in any one equation we get value of a
a=3

Now 8th term of GP=a8=ar⁷
=32⁷
=3x128=384