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

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6, 16 and 29

7, 16 and 25

8, 15 and 25

10, 12 and 26

answer is B.

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Detailed Solution

Given that the sum of the first three terms of an A.P. is 48 and the product of the first and second terms exceeds 4 times the third term by 12.
General Arithmetic Progression:

a − d, a

and

a + d

.
Let the first three terms of the A.P. are:

a − d, a

and

a + d

.
As the sum of the first three terms = 48.

∴ (a − d) + a + (a + d) = 48 ⇒ 3 a = 48 ⇒ a = 483 ⇒ a = 16

As the product of the first and second terms exceeds 4 times the third term by 12.

∴ (a − d) × a = 4 (a + d) + 12

…… (1)
Putting the value of a=16 in equation (1):

∴ (16 − d) × 16 = 4 (16 + d) + 12 ⇒ 256 − 16 d = 64 + 4 d + 12 ⇒ 16 d + 4 d = 256 − 76 ⇒ 20 d = 180

⇒ d = 18020 ⇒ d = ± 9

Take d=9, a= 16, and determining the first three terms of the A.P.:

∴ (16 − 9), 16, and (16 + 9) ⇒ 7, 16, and 25

Take d= -9, a=16, and determining the first three terms of the A.P.:

∴ {16 − (− 9)}, 16, and {16 + (− 9)} ⇒ (16 + 9), 16, and (16 − 9) ⇒ 25, 16, and 7

Therefore, the first three terms of the A.P. are 7, 16 and 25.
Hence, option (2) is correct.

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