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Question
The sum of the first three terms of an A.P. is 33. If the product of the first and the third terms exceeds the second term by 29, find the A.P.
Solution
Let first three terms of A.P. be a – d, a, a + d.
Given, a – d + a + a + d = 33
⇒ 3a = 33
⇒ a = 11 ...(i)
And (a – d)(a + d) = a + 29
⇒ a2 – d2 = a + 29
⇒ (11)2 – d2 = 11 + 29 ...[From equation (i)]
⇒ 121 – d2 = 40
⇒ d2 = 81
⇒ d = ± 9
When, a = 11 and d = 9
Then, A.P. is 2, 11, 20......
When, a = 11 and d = –9
Then, A.P. is 20, 11, 2......
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