Question

if the sum of n terms of an AP IS 2n^2+ 5 , then prove that an= 4n + 3

Answer 1

SOLUTION:
Given:
Sum of the nth term of an AP(Sn) = 2n² + 5n………………..(1)
S(n-1) = 2(n-1)² + 5(n-1)
= 2(n²+1-2n) +5n -5
= 2n² +2 -4n +5n -5
= 2n² -4n +5n +2-5
= 2n² +n -3………………(2)
nth term of an AP (an) = sn - s(n-1)
an = 2n² + 5n -( 2n² +n -3)
[From equation 1 and 2]
an = 2n² + 5n -2n² - n + 3
an = 5n - n +3
an = 4n +3

HOPE THIS WILL HELP YOU...

Answer 2

Solution :-

Sn :- 2n² + 5n

S( n - 1 ) = 2 ( n - 1 )² + 5 (n - 1 )

S ( n - 1 ) = 2n² + 2 - 4n + 5n - 5

S ( n - 1 ) = 2n² - 3 + n

an = Sn - Sn - 1

an = 2n² + 5n - [ 2n² - 3 + n ]

an = 2n² + 5n - 2n² + 3 - n


an = 4n + 3