If the sum of the series 20+1935+1915+1845+ up to the nth term is 488, and the nth term is…

Question

If the sum of the series $20+19\dfrac35+19\dfrac15+18\dfrac45+\cdots$ up to the $n$th term is $488$, and the $n$th term is negative, then:

Accepted Answer

First term $20$, common difference $-\dfrac25$: $S_n = \dfrac{n}{2}\left[\dfrac{200}{5}+(n-1)\left(-\dfrac25\right)\right] = 488 \Rightarrow n^2-101n+2440=0 \Rightarrow n=40$ or $61$.

For $n=40$, $T_n>0$; for $n=61$, $T_n<0$ (the required case).

$$T_{61} = \frac{100}{5}+60\left(-\frac25\right) = 20-24 = -4.$$

Solutions · 1