The sum of the three numbers in an arithmetic progression is 18. If the product of the first and third number is 5 times the common difference, find the three numbers.

Given:

The sum of the three numbers in an arithmetic progression is 18. The product of the first and third number is 5 times the common difference.

To do:

We have to find the three numbers.

Solution:

Let the first three terms of the AP be $a−d,\ a,\ a+d$.

According to the problem,

$a−d+a+a+d=18\ ......( i)$

$( a−d)( a+d)=5d\ .....(ii)$

From $(i)$, we get

$3a=18$

$\Rightarrow a=\frac{18}{3}=6$

From $(ii)$, we get

$a^2−d^2=5d\  .....(iii)$

On putting $a=6$ in equation $(iii)$, we get,

$(6)^2−d^2=5d$

$\Rightarrow d^2+5d-36=0$

$\Rightarrow d^2+9d-4d-36=0$

$\Rightarrow d(d+9)-4(d+9)=0$

$\Rightarrow d+9=0$ or $d-4=0$ 

$\Rightarrow d=-9$ or $d=4$ 

If $a=6, d=-9$, then

$a-d=6-(-9)=6+9=15, a+d=6+(-9)=6-9=-3$

If $a=6, d=4$, then

$a-d=6-4=2, a+d=6+4=10$

The three numbers are $2, 6$ and $10$ or $15, 6$ and $-3$.

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Updated on: 10-Oct-2022

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