Question

The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.

Answer 1

Let the G.P. be a, \(ar,ar^2,ar^3\), …

According to the given condition,

\(a+ar+ar^2=16\space and\space ar^3+ar^4+ar^5\) = 128

⇒ \(a(1+r+r^2)\) = 16 … (1)

\(ar^3 (1 + r + r^2)\) = 128 … (2)

Dividing equation (2) by (1), we obtain

\(\frac{ar^3{(1+r+r^2)}}{a({1+r+r^2)}}=\frac{128}{16}\)

⇒ \(r^3\) = 8

∴ r = 2

Substituting r = 2 in (1), we obtain

a (1 + 2 + 4) = 16

⇒ a (7) = 16

⇒ a = \(\frac{16}{7}\)

\(s_n=\frac{a(r^n-1)}{r-1}\)

⇒ \(s_n\frac{16}{7}\frac{(2^n-1)}{2-1}=\frac{16}{7}{(2^n-1)}\)