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In an A.P. sum of three consecutive terms is 27 and their product is 504. Complete the following activity to find the terms.Activity :Let three consecutive terms be \(a - d\), \(a\), \(a + d\).\(\therefore\ a - d + a + a + d =\) ____\(\therefore\ a =…

Q.3 (A)(i) In an A.P. sum of three consecutive terms is 27 and their product is 504. Complete the following activity to find the terms.

Activity :

Let three consecutive terms be \(a - d\), \(a\), \(a + d\).

\(\therefore\ a - d + a + a + d =\) ____

\(\therefore\ a =\) ____

Similarly :

\((a - d) \times a \times (a + d) =\) ____

\(\therefore\ [(9)^2 - d^2] \times 9 = 504\)

\(\therefore\ (81 - d^2) \times 9 = 504\)

\(\therefore\ d^2 = 81 -\) ____

\(\therefore\ d = \pm 5\)

Thus by putting \(a = 9\) and \(d = 5\) we get three consecutive terms = ____

Or by putting \(a = 9\) and \(d = -5\) we get three consecutive terms = ____

1 answer

Best answer

Answer: \(a = 9\); the blanks are 27, 9, 504, 56; the terms are 4, 9, 14 (for \(d = 5\)) or 14, 9, 4 (for \(d = -5\)).

\(a - d + a + a + d = 27 \Rightarrow 3a = 27 \Rightarrow a = 9\).

\((a - d) \times a \times (a + d) = 504 \Rightarrow (81 - d^2) \times 9 = 504\)

\(81 - d^2 = 56 \Rightarrow d^2 = 25 \Rightarrow d = \pm 5\).

With \(a = 9\), \(d = 5\): the terms are \(9 - 5, 9, 9 + 5\), i.e. 4, 9, 14.

With \(a = 9\), \(d = -5\): the terms are 14, 9, 4.

Check: 4 + 9 + 14 = 27 and \(4 \times 9 \times 14 = 504\) ✓

Doubtly AI · 25 Oct 2026 AI-verified · Oct 2026 Report
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