Question

The sum of first 111 terms of an A.P whose 56th term is 15/37 is:
(a) 40
(b) 35
40) 45
(d) 50​

Answer 1

Step-by-step explanation:

A(56) = a + 55d

15/37 = a + 55d

a + 55d = 15/37 --------(1)

S(111) = 111/2 { 2a + (110)d }

= 111/2 * 2 ( a + 55d )

= 111 * 15/37 = 3*15 = 45