Question

find the sum of gp
√3, 1,1÷√3,.......to 7 elements
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Answer 1

Answer:

The sum of 10 terms of the GP is S_{10}=121(1+\sqrt3)S10=121(1+3)

Step-by-step explanation:

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

To find : The sum of 10 terms of GP?

Solution :

The geometric series is in the form a,ar,ar^2,ar^3,...a,ar,ar2,ar3,...

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

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

First term is a=1

Common ratio is r=\sqrt{3}r=3

The sum formula of GP series is

S_n=\frac{a(1-r^n)}{1-r}Sn=1−ra(1−rn)

Substituting all the values, n=10

S_{10}=\frac{1(1-\sqrt{3}^{10})}{1-\sqrt{3}}S10=1−31(1−310)

S_{10}=\frac{1(1-243)}{1-\sqrt{3}}S10=1−31(1−243)

S_{10}=\frac{-242}{1-\sqrt{3}}S10=1−3−242

Rationalize,

S_{10}=\frac{-242}{1-\sqrt{3}}\times \frac{1+\sqrt3}{1+\sqrt3}S10=1−3−242×1+31+3

S_{10}=\frac{-242(1+\sqrt3)}{1^2-(\sqrt{3})^2}S10=12−(3)2−242(1+3)

S_{10}=\frac{-242(1+\sqrt3)}{1-3}S10=1−3−242(1+3)

S_{10}=\frac{-242(1+\sqrt3)}{-2}S10=−2−242(1+3)

S_{10}=121(1+\sqrt3)S10=121(1+3)

Therefore, The sum of 10 terms of the GP is S_{10}=121(1+\sqrt3)S10=121(1+3)

Answer 2

Answer:

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