Question

the sum of n,2n,3n term of an AP are S,S2,S3 respectively .prove that S3=3(S2-S1)

Answer 1

a = n 

$S_{1}$ = 1/2 ( 2n )

= n 

[tex]S _{2} = 2n + 1d [/tex]

$S_{3} = \frac{3}{2} (2n +2d)$

= 3 ( n+ d)

$S_{3} = 3(S_{2}- S_{1})$

[tex] S_{3} = 3( 2n+1d - n ) [/tex]

$S_{3}= 3 ( n+d)$


so $S_{3} = 3(S_{2}- S_{1})$

Answer 2

Here's the answer. I hope this helps. Mark it as the brainliest, if you want to.