the sum of the first n terms of an AP is given by sn=3n2-n , determine the AP and its - Maths - Arithmetic Progressions

Question

the sum of the first n terms of an AP is given by
sn=3n2-n , determine the AP and its
25thterm

Lalit Mehra · Accepted Answer

Dear Student!

Given, sum of first n terms of an AP, S_n = 3n² – n

S_n = 3n²– n

Replacing n by n – 1, we get

$S_{n - 1} = 3 (n - 1)^{2} - (n - 1)$

Let the n^th term of AP be a_n.

$a_{n} = S_{n} - S_{n - 1} ∴ a_{n} = (3 n^{2} - n) - [3 (n - 1)^{2} - (n - 1)] \Rightarrow a_{n} = 3 [n^{2} - (n - 1)^{2}] - n + (n - 1) \Rightarrow a_{n} = 3 (n^{2} - n^{2} + 2 n - 1) - 1 \Rightarrow a_{n} = 3 (2 n - 1) - 1 = 6 n - 4$

$Now, put n = 1 in a_{n}, we get a = 6 \times 1 - 4 = 2 put n = 2 in a_{n}, we get a_{2} = 6 \times 2 - 4 = 8 put n = 3 in a_{n}, we get a_{3} = 6 \times 3 - 4 = 14 and so on . So, required AP is, 2, 8, 14, . . . . .$

Putting n = 25, we get

a₂₅ = 6 × 25 – 4 = 150 – 4 = 146

Thus, the 25^th term of the given A.P. is 146.

Cheers!

the sum of the first n terms of an AP is given by sn=3n2-n , determine the AP and its 25thterm.

the sum of the first n terms of an AP is given by s_n=3n²-n , determine the AP and its 25^thterm.