If the sum of n terms of an A.P. is cn(n−1) where c=0, then sum of the squares of these terms is

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If the sum of n terms of an A.P. is cn(n - 1) where c 0, then

Question

If the sum of $n$ terms of an A.P. is $c n (n - 1)$ where $c \neq = 0$ , then sum of the squares of these terms is

c^{2} n^{2} (n + 1)^{2}

\frac{2}{3} c^{2} n (n - 1) (2 n - 1)

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

None of these

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Jack Discussed

If the sum of

n

terms of an A.P. is

c n (n - 1)

where

c \neq = 0

, then sum of the squares of these terms is

12 mins ago

Discuss this question LIVE

12 mins ago

Let the

A . P

a, a + d, a + 2 d, \dots, a + (n - 1) d

Given

S_{n} = c n (n - 1)

c n (n - 1) = \frac{n}{2} (2 a + (h - 1) d)

2 c (n - 1) = 2 a + (n - 1) d

2 c = d

- 2 c = 2 a - d

- 2 c = 2 a - 2 c

a = 0

Required series

o, d^{2}, (2 d)^{2}, ........ ((n - 1) d)^{2}

Sum of

(n - 1)

terms square

= o + d^{2} + (2 d)^{2} + ........ + ((n - 1) d)^{2}

= d^{2} [1^{2} + 2^{2} + 3^{2} + \dots + (n - 1)^{2}]

= \sum [(n - 1) \times d]^{2}

= \sum [(n - 1) \times 2 c]^{2}

= 4 c^{2} \times \sum (n - 1)^{2}

= \frac{4 c ^{2} \times ( n - 1 ) n ( 2 n - 1 )}{6}

since sum of square of first n natural numbers

S = \frac{n ( n - 1 ) ( 2 n - 1 )}{6}

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Question Text	If the sum of $n$ terms of an A.P. is $c n (n - 1)$ where $c \neq = 0$ , then sum of the squares of these terms is
Answer Type	Text solution:1
Upvotes	150