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

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8. Sequences and Series

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

A

$5,8,11$

B

$5,8,11$

C

$5,8,11$

D

$5,8,11$

Solution

Let the three numbers in $A.P.$ be $a-d, a,$ and $a+d$

According to the given information,

$(a-d)+(a)+(a+d)=24$ ………$(1)$

$\Rightarrow 3 a=24$

$\therefore a=8$

$(a-d) a(a+d)=440$ ………$(2)$

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

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

$\Rightarrow 64-d^{2}=55$

$\Rightarrow d^{2}=64-55=9$

$\Rightarrow d^{2}=\pm 3$

Therefore, when $d=3,$ the numbers are $5,8$ and $11$ and when $d=-3,$ the numbers are $11,8$ and $5$

Thus, the three numbers are $5,8$ and $11 .$

Standard 11

Mathematics

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