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

Given that, first term of the first AP (a) = 8
and common difference of the first AP (d) = 20
Let the number of terms in first AP be n.
` :'` Sum of first n terms of an AP, `S_(n)=(n)/(2)[2a+(n-1)d]`
` :. S_(n)=(n)/(2)[2xx8+(n-1)20]`
` impliesS_(n)=(n)/(2)(16+20n-20)`
`implies S_(n)=(n)/(2)(20n-4)`
` :. S_(n)=n(10n-2) " " ` ...(i)
Now, first term of the second AP(a') = -30
and common difference of the second AP(d') = 8
` :.` Sum of first 2n terms of second AP,
`impliesS_(2n)=(2n)/(2)[2a'+(2n-1)d'] `
`implies S_(2n)=n[2(-30)+(2n-1)(8)]`
` implies S_(2n)=n[-60+16n-8]`
`implies S_(2n)=n[16n-68] " " ` ...(ii)
Now, by given condition,
Sum of first n terms of the first AP= Sum of first 2n terms of the second AP
`implies S_(n)=S_(2n) " " ` [from Eqs. (i) and (ii)]
`implies n(10n-2)=n(16n-68)`
`implies n[(16n-68)-(10n-2)]=0`
` implies n(16n-68-10n+2)=0`
`implies n(6n-66)=0`
` :. " "n=11 " "[:' n ne 0]`
Hence, the required value of n is 11.