JEE PYQ: Sequence And Series Question 24
Question 24 - 2020 (03 Sep Shift 2)
In the sum of the series $20 + 19\frac{3}{5} + 19\frac{1}{5} + 18\frac{4}{5} + \ldots$ upto $n^{th}$ term is 488 and then $n^{th}$ term is negative, then:
(1) $n = 60$
(2) $n^{th}$ term is $-4$
(3) $n = 41$
(4) $n^{th}$ term is $-4\frac{2}{5}$
Show Answer
Answer: (1)
Solution
$\therefore S_n = 488$
$488 = \frac{n}{2}\left[2\left(\frac{100}{5}\right) + (n-1)\left(-\frac{2}{5}\right)\right]$
$488 = \frac{n}{2}(101-n)$
$\Rightarrow n^2 - 101n + 2440 = 0$
$\Rightarrow n = 61$ or $40$
For $n = 40 \Rightarrow T_n > 0$
For $n = 61 \Rightarrow T_n < 0$
$n^{th}$ term $= T_{61} = \frac{100}{5} + (61-1)\left(-\frac{2}{5}\right) = -4$
BETA
