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2.  If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.

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Let three numbers of AP are a-d, a, a+d.

According to given information ,

a-d+a+a+d=24

3a=24

 \Rightarrow a=8

(a-d)a(a+d)=440

\Rightarrow (8-d)8(8+d)=440

\Rightarrow (8-d)(8+d)=55

\Rightarrow (8^2-d^2)=55

\Rightarrow (64-d^2)=55

\Rightarrow d^2=64-55=9

\Rightarrow d=\pm 3

When d=3, AP= 5,8,11   also if d=-3   ,AP =11,8,5.

Thus, three numbers are 5,8,11.

Posted by

seema garhwal

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