Question

Find three consecutive terms in an arithmetic progression whose sum is 18 and sum of their squares is 140.

Answer 1

Hint: To solve this question first we will assume three consecutive terms in an arithmetic progression as $a-d,a,a+d$. Then we will form equations by using the information given in the question that their sum is 18 and the sum of their squares is 140. Then by simplifying the obtained equations we will get the desired answer.

Complete step by step answer:
We have been given that the sum of three consecutive terms of an A.P. is 18 and sum of their squares is 140.
We have to find the terms.
Now, let us assume that three consecutive terms of an arithmetic progression will be $a-d,a,a+d$, where d is the common difference between the terms.
Now, as given the sum of terms is 18. Then we will get
$\Rightarrow a-d+a+a+d=18$
Now, simplifying the above obtained equation we will get
$\begin{align}
  & \Rightarrow 3a=18 \\
 & \Rightarrow a=\dfrac{18}{3} \\
 & \Rightarrow a=6 \\
\end{align}$
Now, given in the question that the sum of their squares is 140, so we will get
$\Rightarrow {{\left( a-d \right)}^{2}}+{{a}^{2}}+{{\left( a+d \right)}^{2}}=140$
Now, substituting the value $a=6$ in the above equation we will get
$\Rightarrow {{\left( 6-d \right)}^{2}}+{{6}^{2}}+{{\left( 6+d \right)}^{2}}=140$
Now, simplifying the above obtained equation we will get
$\begin{align}
  & \Rightarrow {{6}^{2}}+{{d}^{2}}-2\times 6d+36+{{6}^{2}}+{{d}^{2}}+2\times 6d=140 \\
 & \Rightarrow 36+2{{d}^{2}}-12d+36+36+12d=140 \\
 & \Rightarrow 108+2{{d}^{2}}=140 \\
 & \Rightarrow 2{{d}^{2}}=140-108 \\
 & \Rightarrow 2{{d}^{2}}=32 \\
 & \Rightarrow {{d}^{2}}=\dfrac{32}{2} \\
 & \Rightarrow {{d}^{2}}=16 \\
 & \Rightarrow d=\pm 4 \\
\end{align}$
Since the common difference has two values so we will get two series i.e.
$\Rightarrow 6-\left( -4 \right),6,6+\left( -4 \right)$ and $\Rightarrow 6-4,6,6+4$
$\Rightarrow 10,6,2$ and $\Rightarrow 2,6,10$
Hence above are the required terms of an A.P.

Note: The point to be noted is that in arithmetic progression if the number of terms will be odd then ‘a’ is the middle term of the progression and ‘d’ is the common difference. Always remember to take the terms with the same common difference.