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

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(C) The savings for the first three months are $200, 200, 200$. Total $= 600$.Let the total number of months be $n$. The savings from the $4^{th}$ mon

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$21$

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(C) The savings for the first three months are $200, 200, 200$. Total $= 600$.Let the total number of months be $n$. The savings from the $4^{th}$ month onwards form an Arithmetic Progression $(AP)$ with first term $a = 240$ and common difference $d = 40$.The number of terms in this $AP$ is $(n - 3)$.The sum of savings is given by: $600 + \frac{n-3}{2} [2(240) + (n-3-1)40] = 11040$.$600 + \frac{n-3}{2} [480 + (n-4)40] = 11040$.$600 + (n-3) [240 + (n-4)20] = 11040$.$600 + (n-3) [20n + 160] = 11040$.$(n-3) [20n + 160] = 10440$.$(n-3) [n + 8] = 522$.$n^2 + 5n - 24 = 522$.$n^2 + 5n - 546 = 0$.Solving the quadratic equation: $(n + 26)(n - 21) = 0$.Since $n > 0$,we get $n = 21$ 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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