BITSAT 2019 Paper

(c) a = 25, d = 22 - 25 = -3.
Let n be the no. of terms
Sum= 116; Sum =

n

2

[2a + (n- 1)d]
116 =

n

2

[50 + (n - 1)(- 3)]
or 232 = n[50 - 3n + 3] = n[53 - 3n]
= -3n‌2 + 53 n
⇒ 3n‌2 - 53 + 232 = 0 ⇒ (n-8) (3n-29) = 0
⇒ n = 8 or n =

29

3

, n ≠

29

3

∴ n = 8
∴ Now, T‌8 = a + (8 - 1)d = 25 + 7 x (-3)
= 25 - 21
∴ Last term = 4