Question

find 3 consecutive term which are in AP.whose sum is 24 & product is 440

Answer 1

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Hey there !!

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

▶ 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’ ,

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

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THANKS

#BeBrainly.

Answer 2

5 , 8 , 11 are the terms,

Let the first term be: a-d

Second term : a

Third term: a+d

Thus sum of AP with common difference d is

3 a = 24

a= 8

Now

Product of first and last term

( a-d ) ( a+d ) = 55

a^2 - d^2 = 55

64 -55 = d^2

d^2 = 9

d = +3 or - 3

So the desired terms are 5 , 8 , 11