Question

The sum of multiples of 7 between 0 and 500....is
1.13916
2. 17892
3. 24353
4. 16984
Explain fully....​

Answer 1

We know that Multiplies of 7 are : 7 , 14 , 21 , 28. . . . .so on

Question is to find : Sum of Multiples of 7 between 0 and 500

★  The Last Multiple of 7 between 0 and 500 is 497

So, The Series is : 7 + 14 + 21 + 28 + . . . . . + 497

We can notice that, Above Series is an Arithmetic Progression with :

★  First term (a) = 7

★  Common Difference (d) = Second term - First term = (14 - 7) = 7

Sum of n terms in an Arithmetic Progression is given by :

$\mathsf{\bigstar\;\;S_{n} = \dfrac{n}{2}[2a + (n - 1)d]}$

Where : n is the Number of terms in the Respective Progression

Here, We need to find the Number of terms in the Series

We found that : Last term of the Series is 497

If we find the Position of the number 497, then we can easily say how many terms are in the Series

nth term in an Arithmetic Progression is given by :

★  $\mathsf{T_{n} = a + (n - 1)d}$

$\mathsf{\implies 7 + (n - 1)7 = 497}$

$\mathsf{\implies 7 + 7n - 7 = 497}$

$\mathsf{\implies 7n = 497}$

$\mathsf{\implies n = \dfrac{497}{7}}$

$\mathsf{\implies n = 71}$

It means : Total Number of terms in the Series are 71

Now, Let us find the sum of 71 terms of the Series :

$\mathsf{\implies S_{71} = \dfrac{71}{2}[2(7) + (71 - 1)7]}$

$\mathsf{\implies S_{71} = \dfrac{71}{2}[14 + (70)7]}$

$\mathsf{\implies S_{71} = \dfrac{71}{2}[14 + 490]}$

$\mathsf{\implies S_{71} = \dfrac{71}{2}\times 504}$

$\mathsf{\implies S_{71} = {71}\times 252}$

$\mathsf{\implies S_{71} = 17892}$

Answer : Sum of Multiples of 7 between 0 and 500 is 17892

Answer 2

hey mate here is your:- answer:-

since you have sum of all multiples of 7 that means :- 7,14,21,28................497.

A/Q

here, a=7, d=14-7=7 and L=497, n=71

by using sum of term of AP rule:-

Sn=n/2{a+l}

=71/2{7+497}

=71/2×504

=71×252

=17892 ans

here:- a= first term

n= no. of term

d= common difference

Sn= sum of Nth term

Hope it helps!!