Question

The sum of first 30 and 40 terms of an ap is 2265 and 4020,then find the d of the ap

Answer 1

Sum of n Term of an A.P = (n/2)[2a+(n-1)d]

a - first term
d - common difference
n - number of terms

2265 = (30/2)[2a+(30-1)d]

2265 = 15 [2a+29d]

2265 = 30a + 435d - - - - - - - Eqn(1)


4020 = (40/2)[2a+(40-1)d]

4020 = 20[2a+39d]

4020 = 40a + 780d - - - - - - - Eqn(2)

Eqn(1) X 4 =>

9060 = 120a + 1740d - - - - - - - - Eqn (3)

Eqn(2) x 3 =>

12060 = 120a + 2340d - - - - - - - - - Eqn (4)

Eqn(4) - Eqn(3) =>

3000 = 600d

d = 3000/600 = 5