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The sum of the first three terms of an ap is 33 if the product of first and third excceds the second term by 29 then find the AP?
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The sum of the first three terms of an ap is 33 if the product of firs...
Solution:

Given:
- Sum of first three terms of an AP = 33
- Product of first and third terms exceeds the second term by 29

Let the AP be:
a, a + d, a + 2d

Sum of first three terms:
a + (a + d) + (a + 2d) = 33
3a + 3d = 33
a + d = 11

Product of first and third terms:
a(a + 2d) = (a + d) + 29
a^2 + 2ad = a + 29
Substitute a + d = 11
a^2 + 2ad = 11 + 29
a^2 + 2ad = 40

Substitute a + d = 11:
a^2 + 2a(11) = 40
a^2 + 22a - 40 = 0
(a + 20)(a - 2) = 0
a = -20, 2

Therefore, the AP is:
For a = -20: -20, -9, 2 (not valid as it's not an increasing AP)
For a = 2: 2, 11, 20

So, the AP is 2, 11, 20.
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