How many terms of the A.P. -6, -11/2, -5, …. are needed to give the sum –25?

-25 is the sum of n terms of the given A.P.

The common difference of the A.P. is \( d=-\dfrac{11}{2}+6\)

= \( \dfrac{-11+12}{2}= \dfrac{1}{2}\)

The sum of first n terms of an arithmetic progression is given by the equation

\(s_n= \dfrac{n}{2}[2a+(n-1)d]\)

Substitute S_n=-25, a=-6 and \(d= \dfrac{1}{2}\) in the equation

= \( -25= \dfrac{n}{2}[2(-6)+(n-1)\dfrac{1}{2}]\)

= \( -50= n[-12+\dfrac{n}{2}-\dfrac{1}{2}]\)

= \( -50= n[\dfrac{25}{2}+\dfrac{1}{2}]\)

-100 = n [-25 + n]

= n²-25n+100=0

Factorize the equation.

= n²-5n+20n+100=0

= n(n-5)-20(n-5)=0

= n=20 or 5

Therefore, 5 or 20 terms of the A.P. are needed to give the sum -25  .