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)`.

Answer

N/a

Explanation

Let the first term be 'a' and common difference be 'd' of the A.P.
`S_(1) (n)/(2)[2a + (n - 1)d]" " …(1)`
The sum of 2n terms of A.P.
`S_(2) (2n)/(2)[2a + (2n - 1)d]" " …(2)`
and the sum of 3n terms of the A.P.
`S_(3) (3n)/(2)[2a + (3n - 1)d]" " …(3)`
Now, subtracting equation (1) from (2),
`S_(2) - S_(1) = (2n)/(2) [2a + (2n - 1)d] - (n)/(2)[2a + (n - 1)d]`
`= (n)/(2) [4a + 4nd - 2d - 2a - nd + d]`
`= (n)/(2)[2a + 3nd - d]`
`therefore` `S_(2) - S_(1) = (n)/(2) [2a + (3n - 1)d]`
`rArr` `3(S_(2) - S_(1)) = (3n)/(2) [2a + (3n - 1)d]`
`rArr` `3(S_(2) - S_(1)) = S_(3)" "` [From equation (3)]
Therefore `S_(3) = 3(S_(2) - S_(1))" "` Hence Proved.