Yasmeen saves ₹ 32 during the first month,

Given that,

Yasmeen, during the first month, saves = ₹ 32

During the second month, saves = ₹ 36

During the third month, saves = ₹ 40

Let Yasmeen saves ₹ 2000 during the n months.

Here, we have arithmetic progression 32, 36, 40,…

First term (a) = 32, common difference (d) = 36 – 32 = 4

and she saves total money, i.e., Sn = ₹ 2000

We know that, sum of first n terms of an AP is,

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\Rightarrow$ $2000=\frac{n}{2}[2 \times 32+(n-1) \times 4]$

$\Rightarrow$ $2000=n(32+2 n-2)$

$\Rightarrow \quad 2000=n(30+2 n)$

$\Rightarrow \quad 1000=n(15+n)$

$\Rightarrow \quad 1000=15 n+n^{2}$

$\Rightarrow \quad n^{2}+15 n-1000=0$

$\Rightarrow \quad n^{2}+40 n-25 n-1000=0$

$\Rightarrow \quad n(n+40)-25(n+40)=0 \Rightarrow(n+40)(n-25)=0$

$\therefore \quad n=25 \quad[\because \cap \neq-40]$

Hence, in 25 months will she save ₹ 2000 .

[Since, month cannot be negative]