Question

Find the sum of those integers between 1 and 500 which are multiples of 2 as well as of 5.

Answer 1

Hint: We know that all the numbers between 1 and 500 ending with 0 are the multiples of 2 as well as 5. Because these numbers are completely divided by 2 and 5. These numbers form an Arithmetic Progression (A.P.) series starting from 10 to 490 and the sum of these numbers would be equal to the sum of the series.

Complete step-by-step answer:
Given:
The numbers between 1 and 500 that are the multiples are 2 as well as 5 are given below –
 $ 10,20,30,40,50,...,490 $
Since these numbers form an Arithmetic Progression (A.P.) series.
The first term of the series
 $ a = 10 $
The last term of the series
 $ l = 490 $
And the common difference between two consecutive terms
 $
d = \left( {20 - 10} \right)\\
d = 10
 $
Let us assume the number of the terms in this series is $ n $ .
To find the number of terms we use the formula for the $ n{\rm{th}} $ term. So, we get,
The last term of the series,
 $ l = a + \left( {n - 1} \right)d $
Substituting the values in the formula we get,
 $
\Rightarrow 490 = 10 + \left( {n - 1} \right) \times 10\\
\Rightarrow 480 = \left( {n - 1} \right) \times 10\\
\Rightarrow 48 = n - 1\\
\Rightarrow n = 49
 $
So, there are 49 terms in this series.
Now we have all the values needed to calculate the sum of the series. So, the sum of the series by using the formula for the sum of the Arithmetic Progression $ \left( {{S_n}} \right) $ series is given by –
 $ {S_n} = \dfrac{n}{2}\left( {a + l} \right) $
Substituting the values in the formula we get,
 $
\Rightarrow {S_n} = \dfrac{{49}}{2}\left( {10 + 490} \right)\\
\Rightarrow {S_n} = \dfrac{{49}}{2} \times 500\\
\Rightarrow {S_n} = 12250
 $
Therefore, the sum of the series is 12250.
So, the correct answer is “12250”.

Note: The number of terms need to be calculated first because the formula for the sum of the series requires the number of terms value, so we cannot calculate the sum of the series without the number of terms of the series.