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.
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option 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
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