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 (2009) (a) `(n(4n^2-1)c^2)/6` (b) `(n(4n^2+1)c^2)/3` (c) `(n(4n^2-1)c^2)/3` (d) `(n(4n^2+1)c^2)/6`

If the sum of first n terms of an AP is cn^(2) , then the sum of squares of these n terms is (2009)( a) (n(4n^(2)-1)c^(2))/(6) (b) (n(4n^(2)+1)c^(2))/(3) (c) (n(4n^(2)-1)c^(2))/(3) (d) (n(4n^(2)+1)c^(2))/(6)

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 (2009) (n(4n^2-1)c^2)/6 (b) (n(4n^2+1)c^2)/3 (n(4n^2-1)c^2)/3 (d) (n(4n^2+1)c^2)/6

If the sum of n terms of an A.P.is 2n^(2)+5n then its nth term is 4n-3 b.3n-4 c.4n3 d.3n+4

If the sum of n terms of an A.P. is 2n^2+5n ,\ then its nth term is a. 4n-3 b. 3n-4 c. 4n+3 d. 3n+4

If the sum of n terms of an A.P.be 3n^(2)-n and its common difference is 6, then its first term is 2 b.3 c.1 d.4

Find the sum to n terms of the series whose nth term is 4n^(3)+6n^(2)+2n

Find the sum of n terms of the series (1)/(2*4)+(1)/(4*6)+... (A) (n)/(n+1) (B) (n)/(4(n+1)) (C) (1)/((2n)(2n+2))( D) )(1)/(2^(n)(2^(n)+2))

The sum of n terms of a series is An^(2)+Bn then the n^(th) term is (A) A(2n-1)-B(B)A(1-2n)+B(C)A(1-2n)-B(D)A(2n-1)+B

The sum of the series 1+4+3+6+5+8+ upto n term when n is an even number (n^(2)+n)/(4) 2.(n^(2)+3n)/(2) 3.(n^(2)+1)/(4) 4.(n(n-1))/(4)(n^(2)+3n)/(4)

The sum of the series 1+4+3+6+5+8+ upto n term when n is an even number (n^2+n)/4 2. (n^2+3n)/2 3. (n^2+1)/4 4. (n(n-1))/4 (n^2+3n)/4