Question

find the sum of all multiples of 6 between 200 and 800​

Answer 1

Answer:

Multiples of 6 between 200 and 1100 is 204,210,....,1098.

This form an arithmetic progression.

Where, first term is a=204 and common difference d=6

last term is l=1098

First we find number of terms,

l=a+(n-1)dl=a+(n−1)d

1098=204+(n-1)61098=204+(n−1)6

1098-204=(n-1)61098−204=(n−1)6

894=(n-1)6894=(n−1)6

n-1=\frac{894}{6}n−1=

6

894

n-1=149n−1=149

n=150n=150

Now, The sum of A.P formula is

S_n=\frac{n}{2}[2a+(n-1)d]S

n

=

2

n

[2a+(n−1)d]

S_{150}=\frac{150}{2}[2(204)+(150-1)6]S

150

=

2

150

[2(204)+(150−1)6]

S_{150}=\frac{150}{2}[408+6\times 149]S

150

=

2

150

[408+6×149]

S_{150}=75[408+894]S

150

=75[408+894]

S_{150}=75[1302]S

150

=75[1302]

S_{150}=97650S

150

=97650