Question

31st question’s answer please

Answer 1

Answer:

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 , ....

hope this will help u........