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Find the sum to n terms in the geometric progression&n...

Find the sum to n terms | Class 11 Mathematics Chapter Sequence and Series, Sequence and Series NCERT Solutions

Question:

Find the sum to n terms in the geometric progression $square root of 7 space comma space square root of 21 space comma space 3 square root of 7 space comma space...$

Answer:

The given G.P. is $square root of 7 space comma space square root of 21 space comma space 3 square root of 7 space comma space...$

Here, a = $square root of 7$

$r space equals fraction numerator square root of 21 over denominator square root of 7 end fraction space equals space square root of 3 S subscript n space space end subscript equals space fraction numerator a open parentheses 1 minus r to the power of n close parentheses over denominator 1 minus r end fraction S o S subscript n space equals space fraction numerator square root of 7 space open square brackets 1 minus open parentheses square root of 3 close parentheses to the power of n close square brackets over denominator 1 minus square root of 3 end fraction space space space space space space equals space fraction numerator square root of 7 space open square brackets 1 minus open parentheses square root of 3 close parentheses to the power of n close square brackets over denominator 1 minus square root of 3 end fraction cross times fraction numerator 1 plus square root of 3 over denominator 1 space plus space square root of 3 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space open parentheses B y space r a t i o n a l i z i n g close parentheses space space space space space space space equals space fraction numerator square root of 7 space open parentheses 1 space plus square root of 3 close parentheses open square brackets 1 minus open parentheses square root of 3 close parentheses to the power of n close square brackets over denominator 1 minus 3 end fraction space space space space space space space space equals space fraction numerator minus square root of 7 space open parentheses 1 space plus space square root of 3 close parentheses over denominator 2 end fraction open square brackets 1 minus open parentheses 3 close parentheses to the power of n over 2 end exponent close square brackets space space space space space space space space equals space fraction numerator square root of 7 space open parentheses 1 plus square root of 3 close parentheses over denominator 2 end fraction open square brackets open parentheses 3 close parentheses to the power of n over 2 end exponent minus 1 close square brackets$

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Welcome to the NCERT Solutions for Class 11 Mathematics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 3 , Question 8: Find the sum to n terms in the geometric progression ....

Find the sum to n terms | Class 11 Mathematics Chapter Sequence and Series, Sequence and Series NCERT Solutions

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