Let the sum of n, 2n, 3n terms of an A.P. be `S_1,S_2`and `S_3`, respectively, show that `S_3=3(S_2-S_1)`.

Explanation

Let the first term and common difference of the A.P. be 'a' and 'd' respectively.
`S_(1)` =sum of first 'n' terms
`rArr S_(1)=(n)/(2)[2a+(n-1)d] " " ...(1)`
`S_(2)=` sum of first '2n' terms
`=(2n)/(2)[2a+(2n-1)d]`
`rArr S_(2)=(n)/(2)[4a+(4n-2)d] " " ...(2)`
and `S_(3)` = sum of first '3n' terms
`rArr S_(3)=(3n)/(2)[2a+(3n=1)d] " " ...(3)`
Now, `S_(2)-S_(1)=(n)/(2)[4a+(4n-2)d]-(n)/(2)[2a+(n-1)d]`
`=(n)/(2)[4a+(4n-2)d-2a-(n-1)d]=(n)/(2)[2a+(3n-1)d]`
`rArr 3(S_(2)-S_(1))=(3n)/(2)[2a+(3n-1)d]=S_(3) " " ` Hence Proved.