Question

In the arithmetic sequence $60,56,52,48,....$, starting from the first term, how many terms are needed so that their sum is $368$?

Answer 1

It is given that the sum of the arithmetic sequence $60,56,52,48,....$ is $368$ that is $S_n=368$ where the first term is $a_1=60$, the second term is $a_2=56$

We find the common difference $d$ by subtracting the first term from the second term as shown below:

$d=a_2-a_1=56-60=-4$

We know that the sum of an arithmetic series with first term $a$ and common difference $d$ is $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$, therefore, with $a=60,d=-4$ and $S_n=368$, we have

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]\Rightarrow 368=\frac{n}{2}\left[\right(2\times 60)+(n-1\left)\right(-4\left)\right]\Rightarrow 368\times 2=n(120-4n+4)\Rightarrow 736=n(124-4n)$

$\Rightarrow 736={-4n}^2+124n\Rightarrow 4n^2-124n+736=0\Rightarrow 4n^2-124n+736=0\Rightarrow 4(n^2-31n+184)=0\Rightarrow n^2-31n+184=0$

$\Rightarrow n^2-23n-8n+184=0\Rightarrow n(n-23)-8(n-23)=0\Rightarrow (n-8)(n-23)=0\Rightarrow n=8,n=23$

Hence, thesum of $8$ or $23$ terms of the given arithmetic sequence is $368$.