If 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
Answer: Option B
Solution(By Examveda Team)
$${S_n}$$ = Sum of $$n$$ terms of A.P. and $${S_{2n}}$$ = $$3{S_n}$$$${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right],$$ $${S_{2n}} = \frac{{2n}}{2}\left[ {2a + \left( {2n - 1} \right)d} \right]$$ and $${S_{3n}} = \frac{{3n}}{2}\left[ {2a + \left( {3n - 1} \right)d} \right]$$
We know that $${S_{3n}} = 3\left( {{S_{2n}} - {S_n}} \right)$$ and $${S_{2n}} = 3{S_n}$$
$$\eqalign{ & \Rightarrow \frac{{{S_{3n}}}}{{{S_n}}} = \frac{{3\left( {{S_{2n}} - {S_n}} \right)}}{{{S_n}}} \cr & \Rightarrow \frac{{{S_{3n}}}}{{{S_n}}} = \frac{{3\left( {3{S_n} - {S_n}} \right)}}{{{S_n}}} \cr & \Rightarrow \frac{{{S_{3n}}}}{{{S_n}}} = \frac{{3 \times 2{S_n}}}{{{S_n}}} \cr & \Rightarrow \frac{{{S_{3n}}}}{{{S_n}}} = \frac{6}{1} \cr & \therefore {S_{3n}}:{S_n} = 6 \cr} $$
Related Questions on Progressions
Find the first term of an AP whose 8th and 12th terms are respectively 39 and 59.
A. 5
B. 6
C. 4
D. 3
E. 7
The sum of the first 16 terms of an AP whose first term and third term are 5 and 15 respectively is
A. 600
B. 765
C. 640
D. 680
E. 690
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