Class 11 Mathematics - Chapter Sequence and Series NCERT Solutions | Let the sum of n, 2n, 3n terms

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 Excercise 5 , Question 3: let the sum of nbsp n 2n 3n nbsp terms of an a p....
Question 3

Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively, show that S3 = 3 (S2– S1)

Answer

Let a and b be the first term and the common difference of the A.P. respectively.

Therefore,

S subscript 1 space equals space n over 2 open square brackets 2 a space plus space open parentheses n minus 1 close parentheses d close square brackets space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis
S subscript 2 space equals space fraction numerator 2 n over denominator 2 end fraction open square brackets 2 a plus open parentheses 2 n minus 1 close parentheses d close square brackets space equals space n open square brackets 2 a space plus space open parentheses 2 n minus 1 close parentheses d close square brackets space space... left parenthesis 2 right parenthesis
S subscript 3 space equals space fraction numerator 3 n over denominator 2 end fraction open square brackets 2 a space plus space open parentheses 3 n minus 1 close parentheses d close square brackets space space space space space... left parenthesis 3 right parenthesis

From (1) and (2), we obtain

S subscript 2 space minus space S subscript 1 space equals n open square brackets 2 a space plus space open parentheses 2 n minus 1 close parentheses d close square brackets space minus n over 2 open square brackets 2 a space plus space open parentheses n minus 1 close parentheses d close square brackets
space space space space space space space space space space space space space space space space equals space n open curly brackets fraction numerator 4 a space plus space 4 n d space minus 2 d space minus 2 a space minus n d space plus d over denominator 2 end fraction close curly brackets
space space space space space space space space space space space space space space space space equals space n open square brackets fraction numerator 2 a space plus space 3 n d space minus d over denominator 2 end fraction close square brackets
space space space space space space space space space space space space space space space equals space n over 2 space open square brackets 2 a space plus space open parentheses 3 n minus 1 close parentheses d close square brackets
therefore space 3 open parentheses S subscript 2 space minus space S subscript 1 close parentheses space equals space fraction numerator 3 n over denominator 2 end fraction space open square brackets 2 a space plus space open parentheses 3 n minus 1 close parentheses d close square brackets equals S subscript 3 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 space open square brackets F r o m space left parenthesis 3 right parenthesis close square brackets

Hence, the given result is proved.

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