What will be the nth term of an arithmetic progression if the sum of its n term is given by 2n²+n?

Answer 1

The nth term of this sequence is simply the difference between the sum upto n terms minus the sum upto n-1 terms..

Therefore,

Tn

= 2n²+n - [ 2(n-1)²+(n-1) ]

= 2n²+n - [ 2(n²-2n+1)+(n-1) ]

= 2n²+n - 2n²+4n-2-n+1

= 4n-1 (answer)

PRO TIP: Best to check if the answer holds true by substituting a few simple numbers.

n Tn Sum(n)

1 3 3 (= 2*1^2 + 1)

2 7 10 (=2*2^2 + 2)

3 11 21 (=2*3^2 + 3)

... and so on

Answer 2

I had been working on this problem before Sam Z. answered it.

Respectfully, I do not agree with Sam Z.

Let me start with the answer:

Take a₁=3 and d=4

The sum of an arithmetic series is (n/2)[a₁ + a_n]

In this case: (n/2)[3 + 3+4(n-1)] = 3n + 2n² - 2n = 2n² + n

How is this determined?

(n/2)[a₁ + a₁ + (n-1)d] = 2n² + n

na₁ + n²d/2 - nd/2 = 2n² + n....this equation must be identically true, i.e.

d/2 = 2 => d=4

and a₁ - (4/2) = 1 => a₁ = 3

Answer 3

Nothing more can be done because there is no "like term". n^2 and n are different. You need a value to get a sum.

Answer 4

S₁ = 3

S₂ = 10

d = a₂ - a₁

a₂ = S₂ - S₁

a₂ = 10 - 3

a₂ = 7

a₁ = 3

d = 7 - 3

d = 4

a_n = 3 + (n -1)4

a_n = 3 + 4n - 4

a_n = 4n - 1