Correct Answer - Option 4 : 7
Concept:
If a1, a2, a3,...are in GP with common ratio r, then 1/a1, 1/a2, 1/a3,.......are also in GP with common ratio 1/r.
If a, ar, ar2, ar3,.....,arn-1 are in GP then nth term of GP is given by Tn = arn-1
If a1, a2, a3, a4 are in GP, then common ratio of GP is given by, \(\rm r=\frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}=\frac{a_{4}}{a_{3}}\)
If a be the first term, r be the common ratio of a GP then, \(\rm S_n = \frac{a(1-r^n)}{1-r}=\frac{a}{1-r}-\frac{ar^{n}}{1-r}\)
Let S∞ denote the sum of infinite term of GP, then \(\rm S{_{∞}}= \frac{a}{1-r}\), where -1 < r < 1.
Calculation:
Given GP is 2, 4, 8, 16.....
⇒ a = 2, r = 2 > 1
Let, n terms needed to get the sum 254.
We know that
\(\rm S_n = \frac{a(r^n\ -\ 1)}{r\ -1}\)
\(254 = \frac{2(2^n\ -\ 1)}{2\ -\ 1}\)
⇒ 2n - 1 = 127
⇒ 2n = 27
⇒ n = 7
