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Question

The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the second term by 6, find three terms.


Solution

Let the 3rd term of A.P. be
a - d, a, a + d
Then,
a - d + a + a + d = 21
3a=21
a=7
and
(ad)(a+d)=a+6a2d2=a+6
72a2=7+6        [a=7]
d2=36d=±6
When d = 6, a = 7, we get : 1, 7, 13
When d = -6, a = 7, we get : 13, 7, 1


Mathematics
RD Sharma
Standard XI

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