Question

In a certain geometric progression, the third term exceeds the first term by 9 while the second term exceeds the fourth term by 18. Find the sum of the first four terms.

Answer 1

Solution : -

S(4) = -15

Given : - In G.P.

A(3) = A(1) + 9

A(2) = A(4) + 18

To Find out : -

Sum of first four terms = ?

Explanation : -

Formula :-

[ An = ar^n-1 ]

ar^2 = a + 9

a(r² - 1) = 9 ---------(1)

ar = ar³ + 18

ar³ - ar = -18

ar(r² - 1) = -18 --------(2)

From eq.(2) / eq.(1)

r = -2

then , a = 9/(4 - 1)

a = 9/3 = 3

Now ,

[ Sn = a(r^n - 1)/(r - 1) ]

Sum of first four terms

S(4) = 3{ (-2)⁴ - 1 } / (-2 - 1)

= 3 (15) / (-3)

= 45 / (-3)

= -15