SATHEE: Sequences and Series 2 Question 8

Sequences and Series 2 Question 8

9. If the sum of first $n$ terms of an $A P$ is $c n^{2}$ , then the sum of squares of these $n$ terms is

(a) $\frac{n (4 n^{2} - 1) c^{2}}{6}$

(b) $\frac{n (4 n^{2} + 1) c^{2}}{3}$

(d) $\frac{n (4 n^{2} + 1) c^{2}}{6}$

(2009)

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Answer:

Correct Answer: 9. (b)

Solution:

Let $S_{n} = c n^{2}$

$\begin{array}{cc} S_{n - 1} = c (n - 1)^{2} = c n^{2} + c - 2 c n \\ T_{n} = 2 c n - c [∵ T_{n} = S_{n} - S_{n - 1}] \\ T_{n}^{2} = (2 c n - c)^{2} = 4 c^{2} n^{2} + c^{2} - 4 c^{2} n \end{array}$

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Sequences and Series 2 Question 8

9. If the sum of first n terms of an AP is cn2, then the sum of squares of these n terms is

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9. If the sum of first $n$ terms of an $A P$ is $c n^{2}$ , then the sum of squares of these $n$ terms is