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  • Let Sn denote the sum of the first n terms of an A.P. If S2n = 3Sn then S3n : Sn is equal to A. 4 B. 6 C. 8 D. 10

Let S_n denote the sum of the first n terms of an A.P. If S\textsubscript{2n} = 3S\textsubscript{n} then S\textsubscript{3n} : S\textsubscript{n} is equal to
A. 4
B. 6
C. 8
D. 10

Answers (1)

    Given S_{2n}=3S_{n} \\\\


 \\ S_{n}=\frac{n}{2} \left[ 2a+ \left( n-1 \right) d \right] \\\\ S_{2n}=\frac{2n}{2} \left[ 2a+ \left( 2n-1 \right) d \right] \\\\ S_{2n}=3S_{n} \\\\ n \left[ 2a+ \left( 2n-1 \right) d \right] =\frac{3n}{2} \left[ 2a+ \left( n-1 \right) d \right] \\\\ 4a+4nd-2d=6a+3nd-3d \\\\ 2a=nd+d \\\\


   \\ S_{3n}=\frac{3n}{2} \left[ 2a+ \left( 3n-1 \right) d \right] =\frac{3n}{2} \left[ \left( n+1 \right) d+ \left( 3n-1 \right) d \right] =6n^{2}d \\\\ S_{n}=\frac{n}{2} \left[ 2a+ \left( n-1 \right) d \right] =\frac{n}{2} \left[ \left( n+1 \right) d+ \left( n-1 \right) d \right] =n^{2}d \\\\ \frac{S_{3n}}{S_{n}}=\frac{6n^{2}d}{n^{2}d}=6 \\\\

Hence, the correct option is b.

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infoexpert21

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