The 4th term from the end of an AP -11, -8, -5, …, 49 is

Answer

A, D

Explanation

Here,`a=-11,d=-(-11)=-8+11and t_(n)=49`
`t_(n)=a+(n-1)a` (Formula)
`therefore49=-11+(n-1)xx3...` (Substituting the values)
`therefore49+11=(n-1)xx3`
`therefore(n-1)xx3=49+11`
`therefore (n-1)xx3=60`
`thereforen-1=(60)/(3)`
`thereforen-1=20`
`thereforen=20+1`
`thereforen=21.`
There are 21 terms in this A.P. The fourth term from the end means
`(21-3)=`18th term from the begining i.e. `t_(18)`
`t_(n)=a+(n-1)d` ...(Formula)
`thereforet_(18)=-11+(18-1)xx3` ...(Substituting) the values)
`=-11+17xx3`
`=-11+51 " "thereforet_(18)=40`