JEE MAIN - Mathematics (2011 - No. 17)
A man saves ₹ 200 in each of the first three months of his service. In each of the subsequent months his saving increases by ₹ 40 more than the saving of immediately previous month. His total saving from the start of service will be ₹ 11040 after
19 months
20 months
21 months
18 months
Explanation
Let required number of months $$=n$$
$$\therefore$$ $$200 \times 3 + \left( {240 + 280 + 320 + ...} \right.$$
$$\left. {\,\,\,\,\,\,\,\,\,\,\,\, + {{\left( {n - 3} \right)}^{th}}\,term} \right) = 11040$$
$$ \Rightarrow {{n - 3} \over 2}\left[ {2 \times 240 + \left( {n - 4} \right) \times 40} \right]$$
$$\,\,\,\,\,\,\,\,\,\,\,\, = 11040 - 600$$
$$ \Rightarrow \left( {n - 3} \right)\left[ {240 + 20n - 80} \right] = 10440$$
$$ \Rightarrow \left( {n - 3} \right)\left( {20n + 160} \right) = 10440$$
$$ \Rightarrow \left( {n - 3} \right)\left( {n + 8} \right) = 522$$
$$ \Rightarrow {n^2} + 5n - 546 = 0$$
$$ \Rightarrow \left( {n + 26} \right)\left( {n - 21} \right) = 0$$
$$\therefore$$ $$n = 21$$
$$\therefore$$ $$200 \times 3 + \left( {240 + 280 + 320 + ...} \right.$$
$$\left. {\,\,\,\,\,\,\,\,\,\,\,\, + {{\left( {n - 3} \right)}^{th}}\,term} \right) = 11040$$
$$ \Rightarrow {{n - 3} \over 2}\left[ {2 \times 240 + \left( {n - 4} \right) \times 40} \right]$$
$$\,\,\,\,\,\,\,\,\,\,\,\, = 11040 - 600$$
$$ \Rightarrow \left( {n - 3} \right)\left[ {240 + 20n - 80} \right] = 10440$$
$$ \Rightarrow \left( {n - 3} \right)\left( {20n + 160} \right) = 10440$$
$$ \Rightarrow \left( {n - 3} \right)\left( {n + 8} \right) = 522$$
$$ \Rightarrow {n^2} + 5n - 546 = 0$$
$$ \Rightarrow \left( {n + 26} \right)\left( {n - 21} \right) = 0$$
$$\therefore$$ $$n = 21$$
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