A man saves 200 in each of the first three mo | vedclass

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(C) The savings for the first few months are: Month $1: 200$,Month $2: 200$,Month $3: 200$. From month $4$ onwards,the savings form an Arithmetic Prog

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$21$

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(C) The savings for the first few months are: Month $1: 200$,Month $2: 200$,Month $3: 200$. From month $4$ onwards,the savings form an Arithmetic Progression $(AP)$ with first term $a = 240$ and common difference $d = 40$. Let the total number of months be $n$. The total savings is given by: $600 + \sum_{k=1}^{n-3} [240 + (k-1)40] = 11040$ $600 + \frac{n-3}{2} [2(240) + (n-3-1)40] = 11040$ $600 + (n-3) [240 + (n-4)20] = 11040$ $(n-3) [240 + 20n - 80] = 10440$ $(n-3) [20n + 160] = 10440$ $(n-3)(n+8) = 522$ $n^2 + 5n - 24 = 522$ $n^2 + 5n - 546 = 0$ Solving the quadratic equation $n^2 + 5n - 546 = 0$: $n = \frac{-5 \pm \sqrt{25 - 4(1)(-546)}}{2} = \frac{-5 \pm \sqrt{2209}}{2} = \frac{-5 \pm 47}{2}$ Taking the positive value,$n = \frac{42}{2} = 21$. Thus,the total saving will be $11040$ after $21$ months.

$A$ 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 the immediately previous month. His total saving from the start of service will be $11040$ after ............ months.

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