Question

The sum of three consecutive terms of an A.P. is 18 and their product is 120. Find these terms

Answer 1

Let the three terms be a-d , a , a+d.

Sum of these = 18

∴ 3a = 18

=> a = 6.

Their product = 120.

∴ (a-d)(a)(a+d) = 120

   a³-ad² = 120

   (6)³ - 6d² = 120

   - 6d² = -96

     d² = 16

   ∴ d = 4.

Hence the terms are 2 , 6 , 10.

Answer 2

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GIVEN :

The sum of three terms of an AP = 18

The product of them = 120.

TO FIND :

The terms.

SOLUTION:

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

Where a is the first term of an AP

d is the common difference of an AP.

=> a - d + a + a + d = 18

3a = 18

a = 18/3

a = 6

=> ( a - d) × ( a ) × ( a + d ) = 120

a^3 - a (d^2) = 120

(6) ^3 - 6( d^2) = 120

216 - 120 = 6 (d^2)

96 = 6 (d^2)

16 = d^2

d = 4

The three terms are ,

=> 6 - 4 , 6 , 6 + 4

=> 2 , 6 , 10

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