Question

The sum of $3$ numbers in AP is$18$. If the product of the first and third number is $5$ times the common difference, find the numbers.

Answer 1

Step 1:Find the value of $\mathrm{a}$

Let a be the first term of AP and d be the common difference of given AP.

Let the terms of AP are $\mathrm{a}-\mathrm{d},\mathrm{a},\mathrm{a}+\mathrm{d}$

Given that, the sum of the numbers is $18$.

$\Rightarrow$$(\mathrm{a}-\mathrm{d})+\mathrm{a}+(\mathrm{a}+\mathrm{d})=18$

$\Rightarrow$ $\mathrm{a}-\mathrm{d}+\mathrm{a}+\mathrm{a}+\mathrm{d}=18$

$\Rightarrow$ $\mathrm{a}+\mathrm{a}+\mathrm{a}=18$

$\Rightarrow$ $3\mathrm{a}=18$

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

$\Rightarrow$ $\mathrm{a}=6$

Therefore, the second integer is $\mathrm{a}=6$

Step 2:Find the value of the first and third integer

Let us now consider the second information: the product of the first and third number is $5$ times the common difference.

$\Rightarrow$$(\mathrm{a}-\mathrm{d})\times (\mathrm{a}+\mathrm{d})=5\mathrm{d}$

$\Rightarrow$ ${\mathrm{a}}^2-{\mathrm{d}}^2=5\mathrm{d}$

$\Rightarrow$ $6^2-{\mathrm{d}}^2=5\mathrm{d}$

$\Rightarrow$ $36-{\mathrm{d}}^2=5\mathrm{d}$

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

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

$\Rightarrow$ $\mathrm{d}=4$

Therefore, $\mathrm{d}=4$

As the terms of AP are $\mathrm{a}-\mathrm{d},\mathrm{a},\mathrm{a}+\mathrm{d},$ we get their values as $6-4,6,6+4=2,6,10$.

Hence, the required integers are $2,6,10$.