If the sum of the 8th and 9th terms of an arithmetic progression is 72 and the 4th term is 6, find the common difference

A.
4 
B.
8 
C.
6(frac{2}{3}) 
D.
9(frac{1}{3})
Correct Answer: Option D
Explanation
Let the first term and common difference = a & d respectively.
(T_{n} = text{nth term} = a + (n – 1) d) (A.P)
Given: (T_{4} = 6 implies a + 3d = 6 … (i))
(T_{8} + T_{9} = 72)
(implies a + 7d + a + 8d = 72 implies 2a + 15d = 72 … (ii))
From (i), (a = 6 – 3d)
(therefore) (ii) becomes (2(6 – 3d) + 15d = 72)
(12 – 6d + 15d = 72 implies 9d = 72 + 12 = 84)
(d = frac{84}{9} = 9frac{1}{3})