Question

Find the number of terms of the AP. $64,60,56,....$ having sum $544$.

Answer 1

Step 1. Find the possible values of $n$:

Here, the AP is $64,60,56,....$

The first term, $a=64$ and common difference, $d=64-60=-4$ .

Let $n$ terms of the AP must be taken, then

$S_n=544$

We know that the sum of first $n^{th}$ terms of an AP is, $S_n=\frac{n}{2}\left[2a+(n-1)d\right]$.

$\begin{array}{cc}\Rightarrow & \frac{n}{2}\left[2\times 64+(n-1)\times (-4)\right]=544\\ \Rightarrow & \frac{n}{2}\left[128-4(n-1)\right]=544\\ \Rightarrow & \frac{n}{2}\left[128-4n+4)\right]=544\\ \Rightarrow & n\left[132-4n\right]=1088\\ \Rightarrow & 132n-4n^2=1088\end{array}$

On further simplification,

$\begin{array}{cc}\Rightarrow & 4n^2-132n+1088=0\\ \Rightarrow & n^2-33n+272=0\\ \Rightarrow & n^2-16n-17n+272=0\\ \Rightarrow & n(n-16)-17(n-16)=0\\ \Rightarrow & (n-16)(n-17)=0\end{array}$

$\begin{array}{cc}\Rightarrow & (n-16)=0\text{and}(n-17)=0\\ \Rightarrow & n=16\text{and}n=17\end{array}$

Step 2. Check at which term the sum of the AP is $544$:

Check for $n=16$.

Substitute $a=64,a_n=4,n=16$ in the equation $S_n=\frac{n}{2}\left[a+a_n\right]$ to find the value of $S_{16}$

$\begin{array}{rcl}& \Rightarrow & S_{16}=\frac{16}{2}\left[64+4\right]\\ & =& 8\times 68\\ & =& 544\end{array}$

Check for $n=17$

Substitute $a=64,a_n=4,n=17$ in the equation $S_n=\frac{n}{2}\left[a+a_n\right]$ to find the value of $S_{17}$

$\begin{array}{rcl}& \Rightarrow & S_{17}=\frac{17}{2}\left[64+0\right]\\ & =& \frac{17}{2}\times 64\\ & =& 17\times 32\\ & =& 544\end{array}$

Step 3. Find the value of ${17}^{th}$ term:

Substitute $64$ for $a,-4$ for $d$ and $17$ for $n$ in the equation $a_n=a+(n-1)d$, we get

$\begin{array}{rcl}& \Rightarrow & a_{17}=64+(17-1)(-4)\\ & =& 64-4\times 16\\ & =& 64-64\\ & =& 0\end{array}$

Therefore, the terms may be either $17$ or $16$ both hold true. The double answers are obtained because the ${17}^{th}$ term is zero and it does not affect the net sum.

Hence, the sum of first 16 and 17 numbers will be $544$.