Question

Hey!! Good morning everyone
One more question for you.
If nth term of an AP is 1/3 (2n + 1), then the sum
of its 19 terms is-
(A) 131
(B) 132
(C) 133
(D) 134​

Answer 1

$\huge\bold{\gray{\sf{Answer:}}}$

$\bold{Explanation:}$

$\huge\bold{(c)\: 133}$

Given :

nth term of an AP is 1/3(2n+1)

To Find :

Sum of its 19th term

General form of nth term given by is

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

$\sf= > \dfrac{1}{3} (2n + 1) = a + (n - 1)d$

$\sf= > \dfrac{2n + 1}{3} = a + (n - 1)d$

$\sf= > \dfrac{2n}{3} + \dfrac{1}{3} = nd + (a - d)$

Now , comparing both sides we get,

$\:\:\:\:\:\boxed{\sf d = \dfrac{2}{3}}$

$\sf= > a - d = \dfrac{1}{3}$

$\sf= > a - \dfrac{2}{3} = \dfrac{1}{3}$

$\sf= > a = \dfrac{1}{3} + \dfrac{2}{3} = \dfrac{3}{3} = 1$

$\:\:\:\:\: \boxed{\sf a = 1}$

Now ,we find two values

a=1 and d=2/3

We have to find sum of 19th term

Applying sum formula:

$\boxed{\sf Sn = \dfrac{n}{2}\bigg [2a +\bigg (n - 1\bigg)d\bigg]}$

$\sf S_{19}= \dfrac{19}{2}\bigg [2a +\bigg (n - 1\bigg)d\bigg]$

$\sf => \dfrac{19}{2} \bigg[2 + \bigg(19 - 1\bigg) \dfrac{2}{3} \bigg]$

$\sf = > \dfrac{19}{2} \bigg(2 + 18 \times \dfrac{2}{3} \bigg)$

$\sf = >\bigg( 2 + 12\bigg) \dfrac{19}{2}$

$\sf = > 14 \times \dfrac{19}{2} = 7 \times 19 = 133$

$\boxed{\therefore \sf S_{19} = 133}$