If the sum of the series 20 + 19 3 5 + 19 1 5 + 18 4 5 + . . . . . . . . . . up to n th term is 488 and the n th term is negative, then :

Question

TestHub · Accepted Answer

The given series is an arithmetic progression whose first term is $20$ and common difference is $19 \frac{3}{5} - 20 = - \frac{2}{5}$ .

According to question,

$488 = \frac{n}{2} [2 (20) + (n - 1) (- \frac{2}{5})]$

$\Rightarrow 488 = \frac{n}{2} (40 - \frac{2 n}{5} + \frac{2}{5}) \Rightarrow n^{2} - 101 n + 2440 = 0$

$\Rightarrow n = 61$ or $40$

For $n = 40 \Rightarrow T_{n} = T_{40} = 20 + (40 - 1) (- \frac{2}{5}) = \frac{22}{5} > 0$

For $n = 61 \Rightarrow T_{n} = T_{61} = 20 + (61 - 1) (- \frac{2}{5}) = - 4 < 0$

Thus, $T_{n} = T_{61} = - 4$

Mathematics - Sequence & Series Question with Solution | TestHub