Question

The sum of ‘n’ terms of a progression is (3n² − 5n). Prove that it is an A.Р.

Answer 1

Given: S_n = 3n² − 5n

The n^th term of any progression can be found using the formula:

`a_n = S_n - S_(n - 1)`

Let’s find `S_(n - 1)` by replacing n with (n − 1):

`S_(n - 1) = 3(n - 1)^2 - 5(n - 1)`

= 3(n² − 2n + 1) − 5n + 5

= 3n² − 6n + 3 − 5n + 5

= 3n² − 11n + 8

Now, subtract `S_(n - 1)` from S_n:

a_n = (3n² − 5n) − (3n² − 11n + 8)

a_n = 3n² − 5n − 3n² − 11n − 8

a_n = 6n − 8

An A.P. is defined by a constant common difference (d).

`a_(n - 1)` = 6(n − 1) − 8

= 6n − 6 − 8

= 6n − 14

d = `a_n - a_(n - 1)`

d = (6n − 8) − (6n − 14)

d = 6n − 8 − 6n + 14

d = 6