Question

if the sum of the first n terms of an ap is 3n^2-2n find the ap and find its 19th term

Answer 1

Sum of AP

Sum = n/2 [2a + (n-1)d]
3n^2 - 2n = n/2 [ 2a + (n-1)d]
6n - 4 = 2a + (n-1)d

Thus nth term = 2a + (n-1)d = 6n - 4
  Thus 19th term

A(19) = 6*19 - 4
          = 110
Thus 19th term is 110

Answer 2

Answer:

the A.P. = 1,7,13....  and 109th term is 109

Step-by-step explanation:

Given : $S_n=3n^2-2n$

To Find: the ap and find its 19th term

Solution:

$S_n=3n^2-2n$

Substitute n = 1

$S_1=3(1)^2-2(1)$

$S_1=1$

Thus the first term of A.P. is a = 1

Now substitute n = 2

$S_2=3(2)^2-2(2)$

$S_2=12-4$

$S_2=8$

Thus the sum of first two terms is 8 i.e.

$a_1+a_2=8$

So, $1+a_2=8$

$a_2=7$

So,  $d=a_2-a_1$

$d=7-1$

d =6

So, A.P. = 1,7,13....

Now nth term in A.P. = $a_n=a+(n-1)d$

$a_{19}=1+(19-1)6$

$a_{19}=109$

Hence the A.P. = 1,7,13....  and 109th term is 109