Correct Answer - Option 2 : 6
Given:
The sum of the 5th term and the 15th term of an AP is 116. The sum of the third term and the 7th term of the same AP is 56.
Formula used:
nth term of an AP, Tn = a + (n - 1)d, where ‘a’ is the first term, ‘n’ is the total number of terms and ‘d’ is the common difference.
Calculation:
Let the first term is ‘a’ and common difference is‘d’.
From the question
T5 + T15 = 116
⇒ (a + 4d) + (a + 14d) = 116
⇒ 2a + 18d = 116
⇒ a + 9d = 58 ----(1)
Similarly,
T3 + T7 = 156
⇒ (a + 2d) + (a + 6d) = 56
⇒ 2a + 8d = 56
⇒ a + 4d = 28 ----(2)
By subtracting equation (2) from (1)
5d = 30
⇒ d = 6
∴ Common difference is 6.