Question

: Find five numbers in arithmetic progression whose sum is 25 and the sum of whose squares as 135.

Answer 1

Hint: We have to assume the terms of AP in the question so that the calculated is reduced and the number of variables are equal to the number of equations available.

Complete step-by-step answer:
Before starting to solve the question, we must first know what an AP is. An arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant. Here, we are given that the sum of the terms of AP is 35. Here, we will have to select the terms of AP in such a way that the number of variables are reduced. Thus, we here consider that the third term of AP is a, we assume here that the common difference of the AP is d. Thus the second term of the AP will be\[a\text{ }\text{ }d\] . The first term of the AP will be\[\text{a}2d\] . Similarly the fourth and fifth term of the AP will be \[a+d\] and \[a+2d\] respectively. Thus, we will get the following equation:
$\left( a-2d \right)+\left( a-d \right)+a+\left( a+d \right)+\left( a+2d \right)=25$ .
$\Rightarrow a-2d+a-d+a+a+d+a+2d=25$.
$\Rightarrow 5a=25$ .
$\Rightarrow a=5$ .
Therefore, the third term of AP will be = 5. Now, another information given in question is that the sum of the squares of the individual term of AP is 135. Thus, we get the following equation:
${{\left( a-2d \right)}^{2}}+{{\left( a-d \right)}^{2}}+{{a}^{2}}+{{\left( a+2d \right)}^{2}}+{{\left( a+d \right)}^{2}}=135$.
$\Rightarrow {{\left( 5-2d \right)}^{2}}+{{\left( 5-d \right)}^{2}}+25+{{\left( 5+2d \right)}^{2}}+{{\left( 5+d \right)}^{2}}=135$ .
$\Rightarrow 25+4{{d}^{2}}-20d+25+{{d}^{2}}-2{{d}^{2}}+25+25+4{{d}^{2}}+20d+25+{{d}^{2}}+10d=135$ .
$\Rightarrow 125+10{{d}^{2}}=135$
$\Rightarrow 10{{d}^{2}}=10$
$\Rightarrow {{d}^{2}}=1$
$\Rightarrow d=1$ and $d=-1$ .
Thus, when $d=1$. The AP formed will be:
$5-2\left( 1 \right),5-1\left( 1 \right),5,5+1\left( 1 \right)+5+2\left( 1 \right)$ .
\[5-2,5-1,5,5+1,5+2\] .
$3,4,5,6,7$ .
When $d=-1$ , the AP formed will be:
$5-2\left( -1 \right),5-1\left( -1 \right),5,5+1\left( -1 \right),5+2\left( -1 \right)$ .
$7,5,6,4,3$ .

Note: We can also choose the term as shown: The first term is a, the second term is\[a+d\] ā, the third term is\[a+2d\] , the fourth term is \[a+3d\] and the fifth term is \[a+4d\] . In this case also, the answer would be the same.