If Sn denotes the sum of the first r terms of an A.P. Then, S3n : (S2n – Sn) is
A. n
B. 3n
C. 3
D. None of these
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right], \cr & {S_{2n}} = \frac{{2n}}{2}\left[ {2a + \left( {2n - 1} \right)d} \right]\,{\text{and}} \cr & {S_{3n}} = \frac{{3n}}{2}\left[ {2a + \left( {3n - 1} \right)d} \right] \cr & {\text{Now}}\,{S_{2n}} - {S_n} \cr} $$$$ = \frac{{2n}}{2}\left[ {2a + \left( {2n - 1} \right)d} \right] - $$ $$\frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$
$$ = \frac{n}{2}\left[ {4a + \left( {4n - 2} \right)d} \right] - $$ $$\left[ {2a + \left( {n - 1} \right)d} \right]$$
$$\eqalign{ & = \frac{n}{2}\left[ {4a - 2a + \left( {4n - 2 - n + 1} \right)d} \right] \cr & = \frac{n}{2}\left[ {2a + \left( {3n - 1} \right)d} \right] \cr & = \frac{1}{3}\left( {{S_{3n}}} \right) \cr & \therefore {S_{3n}}:\left( {{S_{2n}} - {S_n}} \right) \cr & = 3:1\,{\text{or}}\,\frac{3}{1} = 3 \cr} $$
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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