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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
\[\text { Let the three terms of the A . P . be }a - d, a, a + d . \]
\[\text { Then, we have }: \]
\[a - d + a + a + d = 21\]
\[ \Rightarrow 3a = 21\]
\[ \Rightarrow a = 7 . . . . (i)\]
\[\text { Also }, (a - d)(a + d) - a = 6\]
\[ \Rightarrow a^2 - d^2 - a = 6\]
\[ \Rightarrow 49 - d^2 - 7 = 6\]
\[ \Rightarrow 36 = d^2 \]
\[ \Rightarrow \pm 6 = d\]
\[\text { When } d = 6, a = 7, \text { we get} : \]
\[1, 7, 13\]
\[\text { When } d = - 6, a = 7, \text{we get }: \]
\[13, 7, 1\]
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