Yasmeen saves Rs 32 during the first month, Rs 36 in the second month and Rs 40 in the third month. If she continues to save in this manner, in how many months will she save Rs 2000?

Given:

Yasmeen saves Rs 32 during the first month, Rs 36 in the second month and Rs 40 in the third month and so on.

To do:

We have to find the time in which her savings will be Rs. 2000.

Solution:

Amount saved during the first month $=Rs.\ 32$

Amount saved during the second month $=Rs.\ 36$

Amount saved during the third month $=Rs.\ 40$

This implies,

Savings raised each successive month $=Rs.\ 36-32=Rs.\ 4$

Let the time in which her savings will be Rs. 2000 be $x$ months.

Amount saved each month for $x$ successive months (in Rupees) is,

$32, 36, ......$(x terms)

This is in A.P., where

First term $a = 32$

Common difference $d = 4$

Number of terms $n = x$

We know that,

Sum of $n$ terms in an A.P. $S_n = \frac{n}{2}[2a + (n – 1) d]$

$S_n = \frac{x}{2}[2 (32) + (x – 1) 4]$

$ 2000= \frac{x}{2}[64 + 4x-4]$

$2(2000) = x(4x+60)$

$4000=4x(x+15)$

$1000=x^2+15x$

$x^2+15x-1000=0$

$x^2+40x-25x-1000=0$

$x(x+40)-25(x+40)=0$

$(x+40)(x-25)=0$

$x+40=0$ or $x-25=0$

$x=25$ or $x=-40$ which is not possible

Therefore, in 25 months her savings will be Rs. 2000.

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