Question

find the sum of all natural numbers which are less than 500 and also which are multiplr of 7

Answer 1

Answer:

Explanation:

Question :

Find the sum of all natural numbers which are less than 500 and also which are multiple of 7

Solution :

${\bold{\red{❖ \: \: \: }}}{\bold{\bf{\underline{According \: to \: the \: question}}}}$

${\bold{\sf{: \: ⟼ \: \: \: \: \: a_{(first \: term)} \: = \: 7}}}$

${\bold{\sf{: \: ⟼ \: \: \: \: \: d_{(common \: difference)} \: = \: 7}}}$

${\bold{\sf{: \: ⟼ \: \: \: \: \: l_{(last \: term)} \: = \: 497}}}$

❐ The given series (7, 14, 21, 28, 35 . . . 497) is in arithmetic progression and we need to find the number of terms

❐ Let's the no. of terms be n, then the equation will be -

${\bold{\rm{: \: ⟼ \: \: \: \: \: a \: + \: (n \: - \: 1)d \: = \: 497}}}$

${\bold{\rm{: \: ⟼ \: \: \: \: \: 7 \: + \: (n \: - \: 1)7 \: = \: 497}}}$

${\bold{\rm{: \: ⟼ \: \: \: \: \: 7 \: + \: 7n \: - \: 7 \: = \: 497}}}$

${\bold{\rm{: \: ⟼ \: \: \: \: \: 7n \: = \: 497}}}$

${\bold{\rm{: \: ⟼ \: \: \: \: \: n \: = \: \frac{497}{7}}}}$

$: \: ⟼ \: \: \: \: \: {\large{\bf{\orange{ \: \: \: \: \: n = 71}}}}$

${\bold{\bf{\pink{Now \: we \: have \: to \: calculate \: the \: sum \: of \: series}}}}$

We know that :

${\large{\red{ \: ✠ \: \: \: \: \:}}}{\bold{\underline{\boxed{\rm{S \: = \: \: \frac{n}{2}(a \: + \: l)}}}}}$

${\bold{\bf{\pink{Putting \: values \: in \: formula : }}}}$

${\bold{: \: ⟼ \: \: \: \: \:}}{\bold{\rm{S \: = \: \frac{71}{2} \: (7 \: + \: 497)}}}$

${\bold{: \: ⟼ \: \: \: \: \:}}{\bold{\rm{S \: = \: \frac{71}{2} \: (504)}}}$

${\bold{: \: ⟼ \: \: \: \: \:}}{\bold{\rm{S \: = \: 71 \: (252)}}}$

${\bold{: \: ⟼ \: \: \: \: \:}}{\large{\bf{\orange{S \: = \: 17892}}}}$

Hence, the sum of all natural numbers which are less than 500 and are multiple of 7 is 17892

Answer 2

Here all natural numbers which are less than 500 and also which are multiple of 7 are

7 , 14 , 21 , 28 , 35 ,...., 497

First term = a = 7

2nd term = 14

3rd term = 21

4th term = 28

2nd term - 1st term = 3rd term - 2nd term

So the numbers from an arithmetic progression

Hence the required sum

=7+14+21+28+35+-------+497

=7(1+2+3+4+5+------+71)

$\sf{ = 7 \times \dfrac{71(71 + 1)}{2} }$

$\sf{ = 7 \times \dfrac{71 \times 72}{2} }$

=7×71×36

=17892