Question

Find the sum of multiple of 8 between 300 and 500

Answer 1

Step-by-step explanation:

1st multiple of 8 after 300 is 304

and multiple of 8 before 500 is 496

a=304, d=8, tn=496

By formula

tn=a+(n-1)d

496=304(n-1)8

496-304=8n-8

192=8n-8

192+8=8n

200=8n

n=200/8

n=25

Sn=n/2(t1+tn)

Sn=25/2×(304+496)

Sn=25/2×800

Sn=25×400

Sn=10000

The sum is 10000

Answer 2

The sum of the multiples of 8 between 300 and 500 is 10,000.

Given:

Multiples of 8 between 300 and 500

To Find:

Find the sum of the multiples of 8 in the given interval.

Solution:

The multiples of 8 will be in the arithmetic mean of the common difference 8.

As we have to find the multiple of 8 in between 300 and 500. The first term would be 304. The last term would be 496.

So, a=304, d=8, and n = 496. We need to find the value of n.

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

$496 = 304 + (n - 1) \times 8$

$192 = 8(n - 1)$

$n - 1 = 24$

$n = 25$

So we have 25 multiples of 8 in between the given interval.

We need to find the sum of the multiples.

The Sum of n terms in an A.P is equal to

$s (n) = \frac{n}{2} (a + l)$

$= \frac{25}{2} (304 + 496)$

$= \frac{25}{2} \times 800$

$= 10000$

Therefore, the sum of the multiples of 8 in between 300 and 500 is 10,000.

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