Question

The sum of first three terms of an AP is 48. If the product of first and second terms exceeds 4 times the third term by 12. Find the AP.

HINT: Let these terms be (a – d), a, (a + d).

Answer 1

Let the first three terms of the AP be (a – d), a and (a + d).

Then, (a – d) + a + (a + d) = 48

⇒ 3a = 48 

⇒ a = 16 

Now, 

(a – d) × a = 4(a + d) + 12   ...(Given) 

⇒ (16 – d) × 16 = 4(16 + d) + 12

⇒ 256 – 16d = 64 + 4d + 12

⇒ 16d + 4d = 256 – 76

⇒ 20d = 180

⇒ d = 9

When a = 16 and d = 9,

a – d = 16 – 9 = 7

a + d = 16 + 9 = 25

Hence, the first three terms of the AP are 7, 16 and 25.