Question

the nth term of an AP is 4n2-1. the 2nd term is...?​

Answer 1

Step-by-step explanation:

Given:

an= 4n^2 -1

We know that,

$\boxed { \:an =a + (n - 1)d}$

when a = 2,

$a2 = {4(2)}^{2} - 1$

$a2 = 16 - 1$

$\boxed{a2 = 15}$

$\mathfrak{ \:Hope \: It \: Helps}$

Answer 2

Answer:

Given, S n

=4n² −n When n=1,

=4(1) 2 −1

=4−1=3

Now, obviously, the sum of the first term will be the first term itself, as there are no other terms involved.

When n=2, S 2

=4(2) 2−1

=16=2=14

The sum of the first 2 terms is equal to a

1 +a 2

Therefore, a 1+a 2=14

We have shown, above, that a 1 =3

Substituting it in a1 +a 2 =14,

3+a 2 =14a =11.

The common difference, d, is the difference between any 2 consecutive terms.

d=a 2 −a 1

=11−3

=8

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