Question

find the sum of the multiples of 8 between 300 and 500

Answer 1

Answer:

The correct answer is $10000$.

Step-by-step explanation:

To find the sum of the multiples of 8 between 300 and 500.

Sum of a series:

The n-th partial sum of a series exists as the totality of the first n terms. The sequence of partial sums of a series sometimes tends to a real limit. If this happens, we say that this limit exists as the sum of the series. If not, we state that the series has no sum. A series can have a sum only if the particular terms tend to zero.

Step 1

Multiples of $8$ - $8,16,24 \ldots$

First multiple of $8$ between  $300$ and $500=304$

Last multiple of $8$ between $300$ and $500=496$

Step 2

Therefore $a=304, l=496$

$&a+(n-1) d=496$

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

$&\Rightarrow(n-1) 8=192$

Therefore, $8 n-8=192$

$8 n=200$

$n=25$

Step 3

Now, the sum of the series

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

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

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

$&=10000$

Therefore, the sum of the series $&=10000$.

#SPJ3

Answer 2

Answer:

The sum of the multiples of 8 between 300 and 500 is 10000.

Step-by-step explanation:

As per the data given in the question,

We have,

Multiples of 8 are= 8,16,24,32...

So, first multiple of 8 greater than 300 is 304

And, last multiple of 8 less than 500 is 496

Therefore,

we need to find the sum of a AP whose first term is 304 and last term is 496 and d is 8.

For that we need to find the number of terms.

$496=a+(n-1)d\\496=304+(n-1)8\\192=(n-1)8\\n-1=192/8\\n-1=24\\n=25$

So, number of terms are 25.

Now to find the sum,

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

where n: no of terms, a: first term, l:  last term

$=\frac{25}{2} (304+496)\\=\frac{25}{2} \times 800\\=25\times 400\\=10000$

Therefore, the sum is 10000.

#SPJ2