Question

The sum of first 9 terms of an AP is 81 and that of first 15 terms is 225.find the sum of its first n terms

Answer 1

Answer:

Step-by-step explanation:

Take a+4d=9 ...................... (1)

Anf a+7d=15........................... (2)

And mark me as brainliest plźzzzzzzzz

Answer 2

Given:

$\text{Sum of the first 9 terms of an AP is 81}$

$\text{Sum of the first 15 terms of an AP is 225}$

To Find:

$\text{Sum of the first n term}$

Formula:

$S_n = \dfrac{n}{2}(2a + (n - 1)d)$

Solution

Form the 1st equation:

$\text{Sum of the first 9 terms of an AP is 81}$

$S_n = \dfrac{n}{2}(2a + (n - 1)d)$

$S_9 = \dfrac{9}{2}(2a + (9 - 1)d)$

$S_9 = \dfrac{9}{2}(2a + 8d)$

$S_9 = 9(a + 4d)$

$S_9 = 9a + 36d$

$9a + 36d = 81$

$a +4d = 9 \text{------------------[ 1 ] }$

Form the 2nd equation:

$\text{Sum of the first 15 terms of an AP is 225}$

$S_n = \dfrac{n}{2}(2a + (n - 1)d)$

$S_{15} = \dfrac{15}{2}(2a + (15 - 1)d)$

$S_{15} = \dfrac{15}{2}(2a + 14d)$

$S_{15} =15(a + 7d)$

$S_{15} =15a + 105d$

$15a + 105d = 225$

$a + 7d = 15 \text{ --------------------- [ 2 ]}$

Equation [2 ] - Equation [ 1 }:

$3d = 6$

$d = 2$

Substitute d = 2 into Equation [ 1 ]:

$a +4(2) = 9$

$a +8 = 9$

$a = 1$

Find the sum of its nth term:

$S_n = \dfrac{n}{2}(2a + (n - 1)d)$

$S_n = \dfrac{n}{2}(2(1) + (n - 1)(2))$

$S_n = \dfrac{n}{2}(2 + 2n - 2)$

$S_n = \dfrac{n}{2}(2n)$

$S_n = n^2$

$\boxed{\textbf{Answer: }S_n = n^2}$