Question

If sum of 45 terms of a arithmetic progression is 3195 find the 23 rd term of an a.p.

Answer 1

HERE IS THE attachment OF THE ANWER

HOPE YOU UNDERSTAND

Answer 2

Given :

• Sum of 45 terms of an arithmetic progression is 3195.

To Find:

• 23rd term of an arithmetic progression

Formula to be used here :

$\large\boxed{S_{n} = \frac{n}{2}[2a + (n - 1)d] }$

where,

• $S_{n}$ is the sum of n terms .
• n is the number of terms .
• a is the first term
• d is the common difference.

Solution:

$S_{45} = \frac {45}{2} [2\times a + ( 45 - 1 )d ]$

$\implies 3195 = \frac{45}{2} [ 2a + 44d]$

$\implies 3195 = \frac {45}{2} \times 2 [ a + 22d ]$

$\implies \frac{3195}{45} = ( a + 22d )$

$\implies 71 = a + 22d$_____Equation 1

$\sf{Now}$,

$\sf{T_{23} = a + ( 23 - 1 )d}$

$\implies \sf{T_{23} = a + 22d}$

Using the above Equation 1 here we get,

$\bf{\sf{ T_{23} = a + 22d = 71}}$