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QJEEM11Maf5e2: JEE Main 2011 Mathematics - Class 11 JEEM11Maf5e2 Solution: A man saves ₹ 200 in each of the first three months of his…

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

Multiple Choice Options

  • A 19 months
  • B 20 months
  • C 21 months
  • D 18 months

Step-by-Step Verified Solution

Let required number of months =n=n

∴\therefore 200×3+(240+280+320+...200 \times 3 + \left( {240 + 280 + 320 + ...} \right.

            +(n−3)th term)=11040\left. {\,\,\,\,\,\,\,\,\,\,\,\, + {{\left( {n - 3} \right)}^{th}}\,term} \right) = 11040

⇒n−32[2×240+(n−4)×40]\Rightarrow {{n - 3} \over 2}\left[ {2 \times 240 + \left( {n - 4} \right) \times 40} \right]

            =11040−600\,\,\,\,\,\,\,\,\,\,\,\, = 11040 - 600

⇒(n−3)[240+20n−80]=10440\Rightarrow \left( {n - 3} \right)\left[ {240 + 20n - 80} \right] = 10440

⇒(n−3)(20n+160)=10440\Rightarrow \left( {n - 3} \right)\left( {20n + 160} \right) = 10440

⇒(n−3)(n+8)=522\Rightarrow \left( {n - 3} \right)\left( {n + 8} \right) = 522

⇒n2+5n−546=0\Rightarrow {n^2} + 5n - 546 = 0

⇒(n+26)(n−21)=0\Rightarrow \left( {n + 26} \right)\left( {n - 21} \right) = 0

∴\therefore n=21n = 21

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