Question

in a geometrical progression the sum of first 3 terms is 14 and the sum of next 3 terms of it is 112 find the geometric progression

Answer 1

Since , the sum of first 3 terms is 14
a + ar + ar^2 = 14
a * ( 1 + r + r^2 ) = 14
1 + r + r^2 = 14/a (1)

Since , the sum of next 3 terms is 112
ar^3 + ar^4 + ar^5 = 112
ar^3 * ( 1 + r + r^2 ) = 112
Now using (1) to substitute value of ( 1 + r + r^2 )
ar^3 * 14/a = 112
r^3 * 14 = 112
r^3 = 112 / 14
r^3 = 8
So,
r = 2

Substituting this value of r in (1)
1 + 2 + 4 = 14 / a
7 * a = 14
a = 2

So, the GP will be :
2 , 4 , 8 , 16 , 32 , ....