If the 7th term of an AP is twice the third term and the sum of the first four terms is 42, find the common difference.
- 6
-
3
- 2
- 1
The correct answer is: B
Explanation
U7 = a + (7 - 1)d= a + 6d
U3 = a + (3 - 1)d
= a + 2d
But U7 = 2(U3)
∴a + 6d = 2(a + 2d)
a + 6d = 2a + 4d
2a - a + 4d - 6d = 0
a - 2d = 0 → eqn1
Sn = n/2 (2a + (n - 1)d)
42 = 4/2 (2a + (4 - 1)d)
42 = 2(2a + 3d)
21 = 2a + 3d → eqn2
eqn1 * eqn2 0 = 2a - 4d
21 = 7d
∴d = 21/7
d = 3