If the sum to n terms of an A.P is cn(n –1); c ≠ 0, then the sum

1.	If the sum to n terms of an A.P is cn(n –1); c ≠ 0, then the sum of squares of these terms is(a) c2 n2 (n + 1)2 (b) 2c2/3 n(n - 1)(n - 1)(c) 2c2/3n(n + 1)(2n + 1)(d) None of thes
Answer» Correct option (b) 2c²/3 n(n - 1)(n - 1) Explanation: T_n = S_n –S_n–1= cn(n–1) –c(n–1)(n–2) = c(n–1){n–n+2} = 2c(n–1) T_n = 4c² (n–1)² ∴ S_n² = 4c²{0²+ 1²+ 2²+..................+ (n – 1)²} = 4c² n(n- 1)(2n - 1)/6 = 2c²/3 n(n - 1)(n - 1)

If the sum to n terms of an A.P is cn(n –1); c ≠ 0, then the sum of squares of these terms is(a) c2 n2 (n + 1)2 (b) 2c2/3 n(n - 1)(n - 1)(c) 2c2/3n(n + 1)(2n + 1)(d) None of thes

Answer»

Correct option (b) 2c²/3 n(n - 1)(n - 1)

Explanation:

T_n = S_n –S_n–1= cn(n–1) –c(n–1)(n–2)

= c(n–1){n–n+2}

= 2c(n–1)

T_n = 4c² (n–1)²

∴ S_n² = 4c²{0²+ 1²+ 2²+..................+ (n – 1)²}

= 4c² n(n- 1)(2n - 1)/6

= 2c²/3 n(n - 1)(n - 1)