Question

find the sum of the first 24 terms of the list of numbers whose nth term is given by an= 3 + 2 n​

Answer 1

${\tt{\green{\underline{\underline{\huge{Question}}}}}}$

Find the sum of the first 24 terms of the list of numbers whose nth term is given by an = 3 + 2n.

$\huge{ { \underbrace{ \mathbb{ \red{Answer\ }}}}}$

Given an = 3 + 2n.

Now,

1st term = a1 = 3 + 2(1)

a = 3 + 2

a = 5

2nd term = a2 = 3 + 2(2)

= 3 + 4

= 7

3rd term = a3 = 3 + 2(3)

= 3 + 6

= 9

4th term = a4 = 3 + 2(4)

= 3 + 8

= 11

So, the series is 5, 7, 9, 11 ............

Since 7 - 5 = 2

9 - 7 = 2

11 - 9 = 2

So, difference of consecutive terms is same.

So, it is an AP.

We need to find sum of first 24 terms.

We use the formula —

Sn = n/2[2a + (n - 1)d]

Where n = 24, a = 5

d = 7 - 5 = 2

Putting these values in formula —

Sum = n/2[2a + (n - 1)d]

= 24/2[2 × 5 + (24 - 1)2]

= 12[10 + (23)(2)]

= 12[10 + 46]

= 12 × 56

= 672

Hence, sum of 24th terms = 672 (Answer).

Explanation:

$\huge \fbox \</strong><strong>p</strong><strong>i</strong><strong>n</strong><strong>k</strong><strong> {Hope \: it \: helps \: you}$