Question

How many terms must be added in an arithmetic sequence whose first term is 78 and whose common difference is -4 to obtain a sum of 702?

Short answer

Number of terms either 27 or 13

Step-by-step solution

Given information 

In an arithmetic sequence,

first term is 78 , a1=78

common difference is -4, d=-4

sum is 702

Sum formula 

$$\begin{gathered} S_n=\frac{n}{2}\left(2a_1+(n-1)d\right) \\ 702=\frac{n}{2}\left(2\left(78\right)+(n-1)(-4)\right) \\ multiplyby2onbothsides \\ 1404=n(156-4n+4) \\ 1404=-4n^2+160n \\ 4n^2-160n+1404=0 \end{gathered}$$

Quadratic formula 

Solve for n using quadratic formula

$$\begin{gathered} 4n^2-160n+1404=0 \\ n=\frac{-\left(-160\right)\pm \sqrt{{\left(-160\right)}^2-4·\hspace{0.17em}4·\hspace{0.17em}1404}}{2·\hspace{0.17em}4} \\ n=\frac{-\left(-160\right)\pm \hspace{0.17em}56}{8} \\ n=\frac{-\left(-160\right)+\hspace{0.17em}56}{8}=27 \\ n=\frac{-\left(-160\right)-\hspace{0.17em}56}{8}=13 \end{gathered}$$

Number of terms either 27 or 13