Question

the least value of n for which 1+2+2^2+....n terms is greater than 100 ( Ans:7)

Answer 1

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Answer 2

The least value of n is 7.

Given:

1+2+2²+...….n terms  > 100

To Find:

The least value of n

Solution:

From the given series,

First-term(a) = 1, Second term(a1) =2 and third term(a3)=2²

Clearly $\frac{a_{1} }{a} =\frac{a_{2} }{a_{1} } =\frac{2}{1}$

The ratio between any two consecutive terms is the Same.
∴ The Given series are in Geometric Progression.

The Sum of n terms in GP is

$\frac{a(r^{n}-1) }{r-1} > 100\\\\ \frac{1(2^{n}-1) }{2-1} > 100\\\\2^{n} -1 > 100\\\\2^{n} > 100+1\\\\2^{n} > 101$

We know that 2^6=64, 2^7=128, and 2^8 =256

The least value of n, for which 2^n >100 is when n=7.

Therefore, the least value of n is 7.

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