Let a be the first term and d be the common difference of the given A.P.,we have a = − d = −

11

2

+ 6 =

−11+12

2

=

1

2

Let n be the number of terms whose sum is – 25.
∴ Sn = – 25
⇒ Sn =

n

2

[2a + (n - 1) d] ⇒ - 25 =

n

2

[−12+(n−1)

1

2

]
⇒ - 50 = - 12n +

n2−n

2

⇒ - 100 = - 24n + n2 - n ⇒ n2 - 25n + 100 = 0
⇒ n2 - 20n - 5n + 100 = 0 ⇒ (n - 5) (n - 20) = 0 ⇒ n = 5 , 20
Hence either 5 terms or 20 terms are needed to give the sum – 25.