A man saves Rs. 200 in each of the first three months of his...

A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent months his saving increases by Rs. 40 more than the saving of immediate preceding month. His total saving from the start of service will be Rs, 11040 after $|\mathbf{2 0 1 1}|$ (A) 19 months (B) 20 months (C) 21 months (D) 18 months

A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent months his saving increases by Rs. 40 more than the saving of immediate preceding month. His total saving from the start of service will be Rs, 11040 after $|\mathbf{2 0 1 1}|$
(A) 19 months
(B) 20 months
(C) 21 months
(D) 18 months

Key Concepts

Arithmetic Sequence

An arithmetic sequence is a progression of numbers where each term after the first is obtained by adding a constant difference. In the context of the problem, the monthly savings after the initial period increase by a constant amount (Rs. 40), making these savings form an arithmetic sequence.

Sum of an Arithmetic Series

The sum of an arithmetic series is calculated using a formula that involves the number of terms, the first term, and the common difference. This formula is essential to compute the total savings over multiple months when the savings increase uniformly each month.

Piecewise Calculation

In problems where the pattern changes over different intervals, the overall calculation is split into parts. Here, the savings are defined for the initial months differently from the subsequent months. The total saving is found by separately computing the fixed savings for the initial period and the arithmetic sum for the later period, then combining them.

Transcript

00:01 So in this question, savings for first three months, savings for first three months, which is equal to 200 per month and 3 into 200, which is equal to 600.

00:32 Okay let let the time taken to save let time taken to save save rs 11 ,040 to be n plus 3, n plus 3 months okay so let time to say rs 11 ,040 to be n plus 3 months so let time take on to save r s 11 ,040 to be n plus 3 months so in here, here we know that here, you know that d is equal to 40 and a is equal to 240.

01:24 A is equal to 240.

01:27 So in, so in n plus three months, n plus three months, three months, sum of n is equal to n by 2 by n by 2a plus n minus 1 into d so from this equation from this equation from this equation the total savings in n plus 3 months is equal to is equal to 600 is the first is the first 3 months savings 600 plus n by 2 2 into n by 2 into 2a is equal to 4 240 a 2a into 2a plus 2a plus n minus 1 d is equal to 40 okay which is equal to the total amount that is 11 ,040 okay now we can reduce 600 from the total value so n minus she will give n minus 2 into 2 into 240 which is equal to 480 plus 40 40 into n 40 minus 40 is equal to 11 ,040 minus 60 which is equal to 11 ,040 minus 600 which is equal to 10 ,000 and 440.

03:33 Okay.

03:36 10 ,000 and 440.

03:38 Now this n by 2 is put to next other side of the equation.

03:47 Then we will get n into 480 minus 40 which is equal to 440 plus 440 plus 40 plus 40 into 10 ,000 440 is equal to 20 ,000880 this is given by this is got by multiplying two in both sides okay now now we can expand the racket so we'll get by expanding and dividing you will get n square n square plus 11 n minus 5 5 .2 is equal to 0...

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