Question

The sum of first n terms of an AP is 3n2-4n. Determine the AP and its 12th term. Also find its nth term.

Answer 1

The value of 12th term is 65.

The value of nth term is 6n-7.

To find:

The value of 12th term of the given series and nth term.

Solution:

The sum of first n term is$3 n^{2}-4 n$

The sum of first n - 1 term is $3(n-1)^{2}-4(n-1)$

The difference of sum of (n – 1) and n will give us the value of $a_{n}$

The value of $a_{n}=6 n-7$

Putting the value of n = 1, 2 to determine the first value of the series and the common difference

$a_{1}=6 \times 1-7 ; a_{1}=-1$

$a_{2}=6 \times 2-7 ; a_{2}=5$

$d=a_{2}-a_{1}=5--1=6$

Now to find the 12th term of series we use the formula of Arithmetic Progression:

$a_{11}=a_{1}+(n-1) d$

$a_{12}=a_{1}+(12-1) d$

$a_{12}=65$

Therefore, the value of twelfth term of series is 65, whereas the value of nth term of series is 6n-7.

Answer 2

Step-by-step explanation:

Is there in picture correct and verified answer