JEE Main 2020 (Online) 3rd September Evening ShiftQuestion 1
If the sum of the series $20+19\frac35+19\frac15+18\frac45+\ldots$ up to $n^{\text{th}}$ term is $488$ and the $n^{\text{th}}$ term is negative, then :
Sequence and Series · Arithmetic Progression
Solution
Formulas used. (1) $S_n=\frac n2\left[2a+(n-1)d\right]$, $\ a_n=a+(n-1)d$Step 1 - The A.P. $20,\ 19.6,\ 19.2,\ 18.8,\dots$: $a=20$, $d=-\tfrac25$. Step 2 - Sum. $$\frac n2\left[40-\frac25(n-1)\right]=488\ \Longrightarrow\ n(101-n)=2440\ \Longrightarrow\ n^2-101n+2440=0 .$$ $$(n-40)(n-61)=0 .$$ Step 3 - Use "the $n$-th term is negative". $$a_{40}=20-\frac25(39)=4.4>0,\qquad a_{61}=20-\frac25(60)=-4<0 .$$ So $n=61$ and the $n$-th term is $-4$. Answer: (D)