Solveeit Logo

Question

Question: An AP has 21 terms. The sum of the three terms in the middle is 129 and of the last three terms is 2...

An AP has 21 terms. The sum of the three terms in the middle is 129 and of the last three terms is 237. Find the AP.

Explanation

Solution

- Hint: The formula for calculating the nth{{n}^{th}} term of an arithmetic progression is as follows
=a+(n1)d=a+(n-1)d
(Where a is the first term of the given arithmetic progression and d is the common difference of the given arithmetic progression)
In this question, we will first add the middle three terms and form an equation with two variables a and d.
Then, we will add the last three terms and so we will make another equation with the same two variables which are a and d.
Now, we will solve the two equations and hence we will get the values of a and d which are the first term and the common difference of the arithmetic progression respectively.

Complete step-by-step solution -

Now, let a and d be the first term and common difference of the given arithmetic progression.
Now, as the sum of the middle three terms is given and we know that the total number of terms in this given arithmetic progression is 21, hence, we can observe that
a10+a11+a12=129{{a}_{10}}+{{a}_{11}}+{{a}_{12}}=129
Now, using the formula given in the hint is as follows

& \left( a\text{ }+\text{ }9d \right)\text{ }+\text{ }\left( a\text{ }+\text{ }10d \right)\text{ }+\text{ }\left( a\text{ }+\text{ }11d \right)\text{ }=\text{ }129 \\\ & 3a+30d=129\ \ \ \ \ ...(a) \\\ \end{aligned}$$ Similarly, we can use this formula to write the sum for the last three terms as well as follows $$\begin{aligned} & {{a}_{21}}+{{a}_{20}}+{{a}_{19}}=237 \\\ & \left( a\text{ }+\text{ 20}d \right)\text{ }+\text{ }\left( a\text{ }+\text{ }19d \right)\text{ }+\text{ }\left( a\text{ }+\text{ }18d \right)\text{ }=\text{ 237} \\\ & 3a+57d=237\ \ \ \ \ ...(b) \\\ \end{aligned}$$ Now, on subtracting equations (a) and (b), we get $$\begin{aligned} & 27d=108 \\\ & d=4 \\\ \end{aligned}$$ Now, on putting this value in (a), we get $$\begin{aligned} & 3a+30\times 4=129\ \\\ & a=3 \\\ \end{aligned}$$ Hence, the value of a=3 and the value of d=4, so, we can write the arithmetic progression as follows $$3,7,11,15......$$ NOTE: - The students can make an error if they don’t know the general term of an arithmetic progression and also the solving of the given system of equations could not have been done otherwise. Also, the students must know how to solve the system of equations as it is also used in the solving of the question as without knowing them, one could never get to the correct answer as here, the elimination method has been used.