Mala saves money every month in a till, starting with $500$ in the first month, $520$ in the second month, $540$ in the third month, and so on. Ramani also saves money in a different till, starting with $150$ in the first month, $190$ in the second month, $230$ in the third month, and so on. We need to find after how many months from the start will the total amount of money saved by both of them be equal.

AlgebraArithmetic ProgressionSum of SeriesEquationsLinear EquationsQuadratic Equations
2025/4/28

1. Problem Description

Mala saves money every month in a till, starting with 500500 in the first month, 520520 in the second month, 540540 in the third month, and so on. Ramani also saves money in a different till, starting with 150150 in the first month, 190190 in the second month, 230230 in the third month, and so on. We need to find after how many months from the start will the total amount of money saved by both of them be equal.

2. Solution Steps

Let nn be the number of months.
Mala's savings form an arithmetic progression with first term a1=500a_1 = 500 and common difference d1=20d_1 = 20. The total savings of Mala after nn months is given by the sum of the arithmetic progression:
SMala=n2[2a1+(n1)d1]S_{Mala} = \frac{n}{2} [2a_1 + (n-1)d_1]
SMala=n2[2(500)+(n1)(20)]S_{Mala} = \frac{n}{2} [2(500) + (n-1)(20)]
SMala=n2[1000+20n20]S_{Mala} = \frac{n}{2} [1000 + 20n - 20]
SMala=n2[980+20n]S_{Mala} = \frac{n}{2} [980 + 20n]
SMala=n(490+10n)S_{Mala} = n(490 + 10n)
SMala=10n2+490nS_{Mala} = 10n^2 + 490n
Ramani's savings form an arithmetic progression with first term b1=150b_1 = 150 and common difference d2=40d_2 = 40. The total savings of Ramani after nn months is given by the sum of the arithmetic progression:
SRamani=n2[2b1+(n1)d2]S_{Ramani} = \frac{n}{2} [2b_1 + (n-1)d_2]
SRamani=n2[2(150)+(n1)(40)]S_{Ramani} = \frac{n}{2} [2(150) + (n-1)(40)]
SRamani=n2[300+40n40]S_{Ramani} = \frac{n}{2} [300 + 40n - 40]
SRamani=n2[260+40n]S_{Ramani} = \frac{n}{2} [260 + 40n]
SRamani=n(130+20n)S_{Ramani} = n(130 + 20n)
SRamani=20n2+130nS_{Ramani} = 20n^2 + 130n
We want to find the number of months nn such that the total savings are equal:
SMala=SRamaniS_{Mala} = S_{Ramani}
10n2+490n=20n2+130n10n^2 + 490n = 20n^2 + 130n
0=10n2360n0 = 10n^2 - 360n
0=10n(n36)0 = 10n(n - 36)
So, n=0n = 0 or n=36n = 36.
Since nn must be greater than 0, n=36n = 36.

3. Final Answer

After 36 months, the total amount of money saved by both of them will be equal.

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