Question

The sum of $5$ numbers in A.P is $30$ and the sum of their square is $220$ .What is the third term of the A.P?

Answer 1

Here we have been given the sum of $5$ numbers of an A.P and the sum of their squares and we have asked the third term of the A.P. So we will use the concept of A.P and let the $5$ numbers such that the difference between the consecutive numbers is same. Then we will add them and put them equal to the sum given. Finally we will solve the value and get our desired answer.

Complete step by step answer:
The sum of $5$ numbers of an A.P is given as $30$ .
Let us take the $5$ numbers as,
$a-2d,a-d,a,a+d,a+2d$
Here $a=$ first term of the A.P and $d=$ Common difference
So as it is given that their sum is $30$ so,
$\Rightarrow a-2d+a-d+a+a+d+a+2d=30$
Take first term value together and common difference value together as follows:
$\Rightarrow \left( a+a+a+a+a \right)+\left( -2d-d+d+2d \right)=30$
Simplifying further we get,
$\Rightarrow 4a+0=30$
$\Rightarrow a=\dfrac{30}{5}$
So we get,
$a=6$
Now we have taken the third term as $a$ .

Hence the third term of the A.P is $6$.

Note: A.P also known as Arithmetic progression is a sequence of a number such that the difference between the consecutive numbers is always the same. That is we add a fixed number to any term to get our next term of the A.P. We have taken the $5$ numbers in such a way that our one unknown variable is cancelled out completely and we get our answer. By using the first term and the common difference we have taken our $5$ numbers usually the general form of an A.P is $a,a+d,a+2d,...a+\left( n-1 \right)d$ but in such type of question we can take the general form as $a-\left( n-1 \right)d+...a-2d,a-d,a,a+d,a+2d,...a+\left( n+1 \right)d$.