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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.
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Solution
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.
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