Question

Q.20

Answer 1

Sum of first n terms=`n/2[2a+(n-1)d]`

`⇒ S_1=n/2[2a+(n-1)d]`

`"Sum of first 2n terms"=2n/2[2a+(2n-1)d]`

`⇒S_2=2n/2[2a+(2n-1)d]`

Sum of first 3n terms=`3n/2[2a+(3n-1)d]`

`⇒S_3=3n/2[2a+(3n-1)d]`

Now,`3(S_2-S_1)=3[2n/2[2a+(2n-1)d]-n/2[2a+(n-1)d]]`

  `= 3n/2[2[2a+(2n-1)d]-[2a+(n-1)d]]` 

`=3n/2[4a+(4n-2)d]-[2a+(n-1)d]]`

`=3n/2[4a-2a+(4n-2-n+1)d]`

`=3n/2[2a+(3n-1)d]`

`=S_3`

`⇒ S_3=3(S_2-S_1)`