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Class 11
Mathematics
Sequence and Series
let the sum of nbsp n 2n 3n nbsp terms of an a p...

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

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

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