Question

Find the sum of all natural numbers between 100 and 1000 which are multiples of 5.

Answer 1

Hint: Here we will proceed by calculating the number of terms i.e. n by using the formula ${a_n} = a + \left( {n - 1} \right)d$. Also we will use the formula of sum of series in A.P. i.e. ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$.

Complete step-by-step answer:
Arithmetic Progression- An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant.
Now we know that the natural numbers lying between 100 and 1000 which are multiples of 5 are 105, 110, …., 995.
Since this sequence forms Arithmetic progression.
Here First term, a = 105
Common difference, d = 5
Using A.P formula,
We get a + (n-1) d = 995
Now we will substitute the values of a and d in the formula,
We get,
105 + (n-1) 5 = 995
Or (n-1) 5 = 995 – 105 = 890
Or n-1 = 178
Or n = 179
$\therefore $ We got n, now we will find sum using the given below formula-
$ \Rightarrow $${S_n}$ = $\dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
Now we will substitute the values into the formula-
$ \Rightarrow {S_n} = \dfrac{{179}}{2}\left[ {2 \times 105 + \left( {179 - 1} \right)5} \right]$

$ = \dfrac{{179}}{2}\left[ {2 \times 105 + 178 \times 5} \right]$
$ = 179\left[ {105 + 445} \right]$
Or ${S_n} = 179 \times 550$
Or ${S_n} = 98450$

Note: While solving this question, we can also use another method like instead of calculating directly the sum of natural numbers i.e.100 to 1000 which are multiples of 5, we can find sum of natural numbers from 100 to 2000 which are multiples of 5 and then subtract the sum of natural numbers above 1000 to 2000 which are multiples of 5. Hence we will get the sum of natural numbers i.e. 100 to 1000 which are multiples of 5.