Question

Find the three consecutive terms which are in AP whose sum is 27 and product is 648

Answer 1

Let the first term and the common difference of the AP be a and d respectively.

Then the numbers will be a, a+d and a+2d

According to question:

a + a + d + a + 2d = 27

3a + 3d = 27

3( a + d ) = 27

a + d = 9

a = 9 - d ....(i)

And a ( a + d )( a + 2d ) = 648

(9 - d)(9 - d + d)( 9 - d + 2d) = 648

9(9 - d)(9 + d) = 648

(9 - d)(9 + d) = 72

9^2 - d^2 = 72

- d^2 = 72-81

d^2 = 9

d = +3 or -3 ( as squares of both

3 and -3 is equal to 9 )

Using d = 3 or d = -3 in (i)

Therefore a = 6 or a = 12

Hence the AP will be 6, 9, 12..... or 12, 9, 6.....

Hope it helps.

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Answer 2

Answer:

a= 6,9,12...... or a= 12,9,6

HOPE IT HELPS YOU