Two Cases, How ?

Relevant wiki: Arithmetic Progressions

Given : $a = 54, d = -3$

$513 = \frac{n}{2} \times (2(54) + (n - 1)-3)$

$\implies 1026 = n (108 - 3n + 3) \implies 1026 = -105n - 3n^2$

$\implies n^2 - 3n + 342 = 0$

$\implies (n -18)(n - 19) = 0 \implies n = 18, 19$

Therefore, the required number of terms are $\color{#20A900}18, 19$

• Bonus question :

The last term that is the 19 th term is $0$ .

$t_{19} = 54 + 18 \times -3 = 54 - 54 = 0$

Therefore, if we have 18 terms or 19 terms the resultant sum will be 513.

Relevant wiki: Arithmetic Progressions

The given series is a arithmetic progression, with the first term $a_1=54$ and common difference $d=-3$ . The sum of the series $S$ is given by:

$\begin{aligned} S & = \frac {n(2a_1+(n-1)d)}2 & \small \color{#3D99F6} \text{where }n \text{ is the number of terms.} \\ 513 & = \frac {n(108-3n+3}2 \\ 1026 & = 111n - 3n^2 \end{aligned}$

$\begin{aligned} \implies 3n^2 - 111n + 1026 & = 0 \\ n^2 - 37n + 342 & = 0 \\ (n-18)(n-19) & = 0 \end{aligned}$

The two numbers of terms $n = \boxed{18, 19}$ . There are two $n$ 's because the 19th term $a_{19} = 54+18(-3) = 0$ .

The numbers are 54 51 48 45 42 39 36 33 30 27 24 21 18 15 12 9 6 3 which are equal to 513. The pattern is minus 3 to the previous term. The number of terms is 18, but they can be 19 if you add 0.