Question

The sum of three number in ap is 27 and the sum of there square is 293 find the number

Answer 1

Answer:

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Answer 2

$\huge{\underline{\underline{\red{\mathfrak{SoLuTiOn :}}}}}$

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➠ Given :

Three numbers are in A. P whose sum is 27 and sum of their squares are 293.

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➠ To Find :

We have to find the three numbers.

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➠ Solution :

Let the three numbers in A. P be (a + d), (a) and (a - d)

So, A. T. Q

$\sf{(a + d) + (a) + (a - d) = 27} \\ \\ \sf{\implies a + \cancel{d} + a + a \cancel{ - d} = 27} \\ \\ \sf{\implies 3a = 27} \\ \\ \sf{\implies a = \frac{\cancel{27}}{\cancel{3}}} \\ \\ \sf{a = 9}$

$\rule{200}{2}$

Now, their squares

$\sf{\implies (a + d)^2 + (a)^2 + (a - d)^2 = 293} \\ \\ \sf{\implies a^2 + d^2 + \cancel{2ad} + a^2 + a^2 + d^2 \cancel{ - 2ad = 293}} \\ \\ \sf{\implies 3a^2 + 2d^2 = 293} \\ \\ \bf{\: \: \: \: \: \: \: \: \: \: \: Putting \: Value \: of \: a } \\ \\ \sf{\implies 3(9)^2 + 2d^2 = 293} \\ \\ \sf{\implies 243 + 2d^2 = 293} \\ \\ \sf{\implies 2d^2 = 293 - 243} \\ \\ \sf{\implies 2d^2 = 50} \\ \\ \sf{\implies d^2 = \frac{\cancel{50}}{\cancel{2}}} \\ \\ \sf{\implies d = \sqrt{25}} \\ \\ \sf{\implies d = \pm 5}$

$\rule{150}{2}$

When d is +5

First Number = a + d = 9 + 5 = 14

Second Number = a = 9

Third Number = a - d = 9 - 5 = 4

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When d is -5

First Number = a + d = 9 + (-5) = 4

Second Number = a = 9

Thor Number = 9 - (-5) = 14