Question

How many term of series 1+3+3^2+3^3+......+3^n-1 must be taken to make the sum equal to 3280?

Answer 1

n=8 is your answer
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Answer 2

Answer:

The number of terms is 8.

Step-by-step explanation:

The sum of a geometric sequence is

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

where, a is first them and r is common ratio.

The given series is

$1+3+3^2+3^3+......+3^{n-1}$

It is a geometric series.

First term = 1

Common ratio = 3

The sum of n terms of this GP is

$S_n=\frac{1(3^n-1)}{3-1}$

$S_n=\frac{(3^n-1)}{2}$

It is given that the sum is 3280.

$3280=\frac{(3^n-1)}{2}$

$6560=(3^n-1)$

$6560+1=3^n$

$6561=3^n$

$3^8=3^n$

On comparing both sides we get

$8=n$

Therefore, the number of terms is 8.