Question

Let the three numbers are in AP such that their sum is $18$ and the sum of their squares is $158$. The greatest number among them is
$\text{1) 10}$
$\text{2) 11}$
$\text{3) 12}$
$\text{4) None of the above}$

Answer 1

Hint:To solve this question we need to have the knowledge of Arithmetic Progression. To solve the problem we will be considering three terms which are $a-d,a,a+d$ . The next step will be to apply all the conditions and find the values for $a$ and $d$, hence finding the three numbers by substituting the values in the variable.

Complete step-by-step solution:
The question asks us to find the value of the greatest number among the three numbers which are in AP when the sum is given as $18$ and the sum of their squares is $158$. To solve this question we will first assume the three numbers to be are $a-d,a,a+d$ where $a-d$ is the first term. Now we will apply the first condition which says that the sum of the three terms is equal to $18$. On writing it mathematically we get:
$\Rightarrow a-d+a+a+d=18$
In the above expression the term “$d$” gets canceled, giving the equation as $3a$ equal to $18$. Its mathematical representation is:
$\Rightarrow 3a=18$
On dividing both the side by $3$ we get:
 $\Rightarrow \dfrac{3a}{3}=\dfrac{18}{3}$
$\Rightarrow a=6$
So, now the three terms become $6-d,6$ and $6+d$. Now we will have to find the value of “$d$”, so for that we will be applying the next condition which says that the sum of squares of the three numbers is equal to $158$. On writing it mathematically we get:
$\Rightarrow {{\left( 6-d \right)}^{2}}+{{6}^{2}}+{{\left( 6+d \right)}^{2}}=158$
To solve the above expression we will apply ${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and ${{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ formulas. On applying the same we get:
$\Rightarrow 36+{{d}^{2}}-12d+36+36+{{d}^{2}}+12d=158$
$\Rightarrow 108+2{{d}^{2}}=158$
$\Rightarrow 2{{d}^{2}}=158-108$
$\Rightarrow {{d}^{2}}=\dfrac{50}{2}$
$\Rightarrow d=\sqrt{25}$
$\Rightarrow d=\pm 5$
Now we got the values of both $a$ and $d$, so on substituting we get the numbers as $6-1,6,6+5$ which are $1,6,11$.
$\therefore $ The greatest number among the three numbers in AP is $\text{2) 11}$.

Note:We would have chosen the three to be x,y,z but considering it as $a-d,a,a+d$ makes the calculation easier. When this type of question is given with five numbers, then the numbers considered are $a-2d,a-d,a,a+d,a+2d$.