Question

find the 3 consecutive terms in ap whose sum is 18 and sum of their squares is 140

Answer 1

For any AP, if n is odd,

$S_{n} = n * [\frac{n + 1}{2}]^{th} term$

($S_{n}$ means sum of any n consecutive terms of the sequence, not first n consecutive terms.)

Here,

$S_{3} = 3 * [\frac{3 + 1}{2}]^{th} term \\ \\ = 3 * T_{2} = 18 \\ \\ T_{2} = \frac{18}{3} = 6$

∴ The middle term of the 3 consecutive terms is 6.

6² = 36.

140 - 36 = 104 is the sum of the squares of the first and the third terms.

Let the first and the third terms be (6 - d) and (6 + d) respectively. (d = common difference)

$(6 - d)^2 + (6 + d)^2 = 104 \\ \\ = 36 - 12d + d^2 + 36 + 12d + d^2 = 104 \\ \\ = 72 + 2d^2 = 104 \\ \\ 2d^2 = 104 - 72 = 32 \\ \\ d^2 = \frac{32}{2} = 16 \\ \\ d = \sqrt{16} = 4$

∴ 4 is the common difference.

$T_{1} = 6 - 4 = 2 \\ \\ T_{3} = 6 + 4 = 10$

∴ The numbers are 2, 6, and 10.

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Answer 2

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