The sum of the first n terms of an AP whose first term is `8` and the common difference is `20` is equal to the sum of first `2n` terms of another AP whose first term is `-30` and the common difference is `8`. Find `n`.

Explanation

For first A.P., a=8, d=20
`:. S_(n)=(n)/(2)[2(8)+(n-1)(20)]`
`=n(8+10n-10)=10n^(2)-2n`
For second A.P., a=-30, d=8
`S_(2n)=(2n)/(2)[2(-30)+(2n-1)(8)]`
`=n(-60+16n-8)=16n^(2)-68n`
Given that, `16n^(2)-68n=10n^(2)-2n`
`rArr 6n^(2)-66n=0`
`rArr 6n(n-11)=0`
`rArr n=0 or n=11`.
but n=0 is not possible
`:. n=11`