Question

find the sum of first 40 multiple of 8​

Answer 1

Step-by-step explanation:

Let S denote the given sum.

Thus, S = 8+16+24+...+312+320

This is an A.P. with first term being 8 and common difference of 8, having 40 terms.

We know that sum of n terms of an A.P. with first term as a and common difference as d is given by

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

Here, n=40, a=8, d=8.

Thus, S = 40/2*[2*8 + 39*8]

S = 20*(16 + 312)

S = 20*(328)

S = 6560

Thus, required sum is 6560.

Answer 2

QUESTION:

find the sum of first 40 multiple of 8

☆FORMULA USED :

$\huge\blue {s(n) = \frac{n}{2} [2a + (n - 1)d]}$

where;

a = first term

d = common difference

Now,

Multiples of 8 are;

$\red {8...16...24...32....40.....48...56.....}$

According to formula here;

a = 8

d = 16-8

d = 8

n = 40

¤putting the values in the formula;

$\pink {s(40) = \frac{40}{2}[2 \times 8 + (40 - 1)8]}$

$s(40) = 20(16 + 39 \times 8)$

$s(40) = 20 \times 328$

$s(40) = 6560$

FINAL ANSWER :

$\huge\purple {s(40) = 6560}$

☆Some other formula related to AP.

1.

$nth \: term = a + (n - 1)d$

where;

a = first term

d = common difference

2.

$last \: term = l - (n - 1)d$