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(B) Total savings in the first $3$ months $= 3 	imes 200 = 600$ rupees.After $3$ months,the savings form an arithmetic progression: $240, 280, 320, \

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$21$

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(B) Total savings in the first $3$ months $= 3 	imes 200 = 600$ rupees.After $3$ months,the savings form an arithmetic progression: $240, 280, 320, \dots$The sum of these terms must be $11040 - 600 = 10440$ rupees.Here,$a = 240$ and $d = 40$.Using the sum formula $S_n = \frac{n}{2} [2a + (n - 1)d]$:$10440 = \frac{n}{2} [2(240) + (n - 1)40]$$10440 = n(240 + 20n - 20)$$10440 = 220n + 20n^2$$20n^2 + 220n - 10440 = 0$Dividing by $20$: $n^2 + 11n - 522 = 0$$(n + 29)(n - 18) = 0$Since $n > 0$,$n = 18$.Total months $= 18 + 3 = 21$ months.

$A$ man saves $200$ rupees in each of the first three months of his job. In the subsequent months,his savings increase by $40$ rupees compared to the previous month. After how many months from the start of the job will his total savings be $11040$ rupees?

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