The sum of first n terms of a certain series is given as 2n2 – 3n. S

1.	The sum of first n terms of a certain series is given as 2n2 – 3n. Show that the series is an A.P.
Answer» Let t_n be n^th term of an A.P. t_n = S_n – S_{n - 1} = 2n² – 3n – [2(n – 1)² – 3(n – 1)] = 2n² – 3n – [2(n² – 2n + 1) – 3n + 3] = 2n² – 3n – [2n² – 4n + 2 – 3n + 3] = 2n² – 3n – [2n² – 7n + 5] = 2n² – 3n – 2n² + 7n – 5 t_n = 4n – 5 t₁ = 4(1) – 5 = 4 – 5 = -1 t₂ = 4(2) -5 = 8 – 5 = 3 t₃ = 4(3) – 5 = 12 – 5 = 7 t₄ = 4(4) – 5 = 16 – 5 = 11 The A.P. is -1, 3, 7, 11,… The common difference is 4 ∴ The series is an A.P.

The sum of first n terms of a certain series is given as 2n2 – 3n. Show that the series is an A.P.

Answer»

Let t_n be n^th term of an A.P.

t_n = S_n – S_{n - 1}

= 2n² – 3n – [2(n – 1)² – 3(n – 1)]

= 2n² – 3n – [2(n² – 2n + 1) – 3n + 3]

= 2n² – 3n – [2n² – 4n + 2 – 3n + 3]

= 2n² – 3n – [2n² – 7n + 5]

= 2n² – 3n – 2n² + 7n – 5

t_n = 4n – 5

t₁ = 4(1) – 5 = 4 – 5 = -1

t₂ = 4(2) -5 = 8 – 5 = 3

t₃ = 4(3) – 5 = 12 – 5 = 7

t₄ = 4(4) – 5 = 16 – 5 = 11

The A.P. is -1, 3, 7, 11,…

The common difference is 4

∴ The series is an A.P.