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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.
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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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