Question

8. First term of an arithmetic progression is 8, nth
term is 33 and sum of first n terms is 123, then
find n and common difference d.
​

Answer 1

$\huge\mathfrak{Answer:}$

Given:

• We have been given that first term of arithmetic progression is 8, nth term is 33 and sum of its first n terms is 123.

To Find:

• We need to find the number of terms and the common difference.

Solution:

Given,

$\sf{a = 8}$

$\sf{an = 33}$

$\sf{sn = 123}$

We know that,

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

Substituting the values, we have

$\implies\sf{ 33 = 8 + (n - 1)d}$

$\implies\sf{ 33 - 8 = (n - 1)d}$

$\implies\sf{ 25 = (n - 1)d}$

Now, we have

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

Substituting the values, we have

$\sf{123 = \dfrac{n}{2} [2 \times 8 + 25]}$

$\implies\sf{ 123 \times 2 = n(2 \times 8 + 25)}$

$\implies\sf{ 246 =n( 16 + 25)}$

$\implies\sf{246 = n(41)}$

$\implies\sf{ \dfrac{246}{41} = n}$

$\implies\sf{ n = 6}$

Now, we can find the value of d by substituting the value of n in n(n - 1)d = 25.

$\implies\sf{ (6 - 1)d = 25}$

$\implies\sf{ (5)d = 25}$

$\implies\sf{ d = \dfrac{25}{5}}$

$\implies\sf{ d = 5}$

.Hence, the value of n is 6 and the value of d is 5.

Answer 2

Answer:

Not understanding.......