Question

Find the sum of 10 terms of GP series 1 + √3 + 3

Answer 1

Give G.P series,
1+√3+3.................
common ratio=a2/a1=√3/1
common ratio=√3
sum of n terms=Sn={a(1-r^n)}/(1-r)
Here,
a=1 , r=√3 and n=10
Now we have,
S10={1(1-√3^10)}/(1-√3)

S10={1-243}/(1-√3)

S10={-242(1+√3)}/(1-√3)(1+√3)

S10={-242(1+√3)}/(-2)

S10=121(1+√3)

Hence sum of 10 terms of given G.P series=121(1+√3).

Answer 2

Answer:

The sum of 10 terms of the GP is $S_{10}=121(1+\sqrt3)$

Step-by-step explanation:

Given : GP series $1+\sqrt3+3+...$

To find : The sum of 10 terms of GP?

Solution :

The geometric series is in the form $a,ar,ar^2,ar^3,...$

Where, a is the first term and r is the common ration.

In the given, GP series $1+\sqrt3+3+...$

First term is a=1

Common ratio is $r=\sqrt{3}$

The sum formula of GP series is

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

Substituting all the values, n=10

$S_{10}=\frac{1(1-\sqrt{3}^{10})}{1-\sqrt{3}}$

$S_{10}=\frac{1(1-243)}{1-\sqrt{3}}$

$S_{10}=\frac{-242}{1-\sqrt{3}}$

Rationalize,

$S_{10}=\frac{-242}{1-\sqrt{3}}\times \frac{1+\sqrt3}{1+\sqrt3}$

$S_{10}=\frac{-242(1+\sqrt3)}{1^2-(\sqrt{3})^2}$

$S_{10}=\frac{-242(1+\sqrt3)}{1-3}$

$S_{10}=\frac{-242(1+\sqrt3)}{-2}$

$S_{10}=121(1+\sqrt3)$

Therefore, The sum of 10 terms of the GP is $S_{10}=121(1+\sqrt3)$