Question

How many terms of the AP: 15, 13, 11, ___ are needed to make the sum 55? Explain the reason for the double answer.

Answer 1

Hint: For solving this problem, first identify the first term and the common difference of the given arithmetic progression. Let the sum of the n terms of AP be 55. By using the formula for the sum of a series in AP, we obtain the number of terms.

Complete step-by-step solution -
According to the problem statement, we are given the AP: 15, 13, 11, __. Hence, the first term of the series in AP is 15. Now, the common difference of the AP is the difference between the first and second term.
$\begin{align}
\Rightarrow & d={{a}_{2}}-{{a}_{1}} \\
\Rightarrow & d=13-15 \\
\Rightarrow & d=-2 \\
\end{align}$
We are also given that the sum of the AP is 55. Let n terms be used to make the sum 55.
The sum of a series in AP is defined as: ${{S}_{n}}=\dfrac{n}{2}\left\{ 2a+\left( n-1 \right)d \right\}$
Now, putting a = 15, d = -2 and S = 55, we get
\[\begin{align}
 \Rightarrow & 55=\dfrac{n}{2}\left\{ 2\times 15+\left( n-1 \right)\times \left( -2 \right) \right\} \\
\Rightarrow & 110=n\left( 30-2n+2 \right) \\
\Rightarrow & n\left( 32-2n \right)=110 \\
\Rightarrow & \dfrac{32n-2{{n}^{2}}}{2}=\dfrac{110}{2} \\
\Rightarrow & 16n-{{n}^{2}}=55 \\
\Rightarrow & {{n}^{2}}-16n+55=0 \\
\end{align}\]
By using splitting the middle term method, we get
$\begin{align}
\Rightarrow & {{n}^{2}}-5n-11n+55=0 \\
\Rightarrow & n\left( n-5 \right)-11\left( n-5 \right)=0 \\
\Rightarrow & \left( n-11 \right)\left( n-5 \right)=0 \\
\Rightarrow & n=11,5 \\
\end{align}$
Therefore, the number of terms needed to make the sum 55 are 5 or 11.
There exist two answers for this problem. The reason is explained in the note.

Note: Students must not get confused by seeing two possible answers for the given situation. Both of the answers are correct because the AP is decreasing having some initial positive numbers. So, we can get 55 by adding only positive numbers and another 55 by adding positive numbers up to 0 with some of the negative numbers.