Question

find the sum of first 10 terms of the gp 1,root 3,3,3root3​

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).

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Step-by-step explanation:

Answer 2

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

Step-by-step explanation:

Given : The G.P is $1,\sqrt{3},3,3\sqrt{3},....$

To find : The sum of first 10 terms of the G.P ?

Solution :

The G.P is  $1,\sqrt{3},3,3\sqrt{3},....$

The first term is a=1

The common ratio is $r=\frac{\sqrt3}{1}=\sqrt{3}$

Here, n=10

The sum of n terms of G.P is

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

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

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

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

Rationalize by multiplying and divide by $\sqrt{3}+1$

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

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

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

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

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

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

#Learn more

The first term of a gp is 27 and its 8 term is 1 by 81 find the sum of its first 10 terms.

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