If the sum of the series
20 + 19${3 \over 5}$ + 19${1 \over 5}$ + 18${4 \over 5}$ + ...
upto nth term is 488
and the nth term is negative, then :
Solution
$S = {{100} \over 5} + {{98} \over 5} + {{96} \over 5} + {{94} \over 5} + ...\,n$<br><br>$$\,{S_n} = {n \over 2}\left( {2 \times {{100} \over 5} + (n - 1)\left( {{{ - 2} \over 5}} \right)} \right) = 488$$<br><br>$\Rightarrow$ $n(100 - n + 1) = 488 \times 5$<br><br>$\Rightarrow$ ${n^2} - 101n + 488 \times 5 = 0$<br><br>$\Rightarrow$ $n = 61,\,40$
<br><br>For negative term n = 61
<br><br>$\,{T_n} = a + (n - 1)d = {{100} \over 5} - {2 \over 5} \times 60$<br><br>$= 20 - 24 = - 4$
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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