Class 11 Mathematics - Chapter Sequence and Series NCERT Solutions | Let S be the sum, P the product and R th

Welcome to the NCERT Solutions for Class 11th Mathematics - Chapter Sequence and Series. This page offers a step-by-step solution to the specific question from Exercise 5, Question 14: let s be the sum p the product and r the sum of r....
Question 14

Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that P2Rn = Sn

Answer

Let the G.P. be aarar2ar3, … arn – 1

According to the given information,

S space equals space fraction numerator a space open parentheses r to the power of n space minus 1 close parentheses over denominator r space minus 1 end fraction
P space equals space a to the power of n space end exponent cross times space r to the power of 1 plus 2 plus... plus n minus 1 end exponent
space space space space equals space a to the power of n space end exponent r to the power of fraction numerator n open parentheses n minus 1 close parentheses over denominator 2 end fraction space space space space space space space open square brackets because space S u m space o f space f i r s t space n space n a t u r a l space n u m b e r s space i s space fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 end fraction close square brackets end exponent

R space equals space 1 over a space plus space fraction numerator 1 over denominator a r end fraction space plus space... space plus space fraction numerator 1 over denominator a r to the power of n minus 1 end exponent end fraction
space space space space equals space fraction numerator r to the power of n minus 1 end exponent plus r to the power of n minus 2 end exponent space plus space... r plus 1 over denominator a r to the power of n minus 1 end exponent end fraction
space space space space equals space fraction numerator 1 open parentheses r to the power of n space minus 1 close parentheses over denominator open parentheses r minus 1 close parentheses end fraction space cross times space fraction numerator 1 over denominator a r to the power of n minus 1 end exponent end fraction space space space space space space space space space space open square brackets because space 1 comma r comma... r to the power of n minus 1 end exponent space f o r m s space a space G. P. close square brackets
space space space space equals space fraction numerator r to the power of n space minus 1 over denominator a r to the power of n minus 1 end exponent open parentheses r minus 1 close parentheses end fraction
therefore space P squared R to the power of n space equals space a to the power of 2 n end exponent r to the power of n open parentheses n minus 1 close parentheses space end exponent fraction numerator open parentheses r to the power of n space minus 1 close parentheses to the power of n over denominator a to the power of n r to the power of n open parentheses n minus 1 close parentheses end exponent open parentheses r minus 1 close parentheses to the power of n end fraction
space space space space space space space space space space space space space space space space space equals space fraction numerator a to the power of n open parentheses r to the power of n minus 1 close parentheses to the power of n over denominator open parentheses r minus 1 close parentheses to the power of n end fraction
space space space space space space space space space space space space space space space space space equals space open square brackets fraction numerator a open parentheses r to the power of n space minus 1 close parentheses over denominator open parentheses r minus 1 close parentheses end fraction close square brackets to the power of n
space space space space space space space space space space space space space space space space equals space S to the power of n

Hence P2 Rn= Sn

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