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Question

Find three consecutive terms which are in AP . Whose sum is 24 and product is 440

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Solution

→ Let the required number be ( a- d ) , a and ( a + d ).....................(i)

▶ Now,

A/Q,

=> ( a - d ) + a + ( a + d ) = 24.


=> a -d + a + a + d = 24.

=> 3a = 24.

=> a = 24/3.

=> a = 8.

▶ Now, again

A/Q,

=> ( a - d ) × a × ( a + d ) = 440.

=> a( a² - d² ) = 440.

=> 8 ( 8² - d² ) = 440.

=> 8 ( 64 - d² ) = 440.

=> 512 - 8d² = 440.

=> 8d² = 512 - 440.

=> 8d² = 72.

=> d² = 72/8.

=> d = √9.

=> d = ±3.

By putting the value of ‘a’ and ‘d’ in equation (i) we get the required Number.

Hence, the required numbers are ( 5,8,11 ) and ( 11,8,5 ) .

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