Question

How many terms must be added in an arithmetic sequence whose first term is 11 and whose common difference is 3 to obtain a sum of 1092?

Short answer

24 terms added to obtain a sum of 1092

Step-by-step solution

Given information 

first term is 11 , a1=11

common difference is 3 , d=3

sum is 1092

sum formula

Sum of n terms formula is

$$\begin{gathered} S_n=\frac{n}{2}\left(2a_1+(n-1)d\right) \\ substituteallthevaluesinsidetheformula \\ 1092=\frac{n}{2}\left(2\left(11\right)+(n-1)3\right) \\ solveforn \\ 1092=\frac{n}{2}\left(22+3n-3\right) \\ multiplyby2onbothsides \\ 2184=n(3n+19) \\ 2184=3n^2+19n \\ 3n^2+19n-2184=0 \end{gathered}$$

solve for n 

$$\begin{gathered} applyquadraticformulatosolveforn \\ n=\frac{-19\pm \sqrt{{19}^2-4·\hspace{0.17em}3\left(-2184\right)}}{2·\hspace{0.17em}3} \\ n=\frac{-19\pm \hspace{0.17em}163}{6} \\ n=\frac{-19+\hspace{0.17em}163}{6}=24 \\ n=\frac{-19-\hspace{0.17em}163}{6}=-\frac{91}{3} \end{gathered}$$

number of terms cannot be negative

So, n= 24

24 terms added to obtain a sum of 1092