Given that,
Yasmeen, during the first month, saves = Rs. 32
During the second month, saves = Rs. 36
During the third month, saves = Rs. 40
Let Yasmeen saves Rs. 2000 in "n" months.
Here, we have an arithmetic progression 32, 36, 40……
First term (a) = 32, common difference (d) = 36 - 32 = 4
Sn = Rs. 2000
We know that, sum of first n terms of an AP is,
Sn=n2[2a+(n−1)d]
⇒2000=n2[2×32+(n−1)×4]
⇒2000=n(32+2n−2)
⇒2000=n(30+2n)
⇒1000=n(15+n)
⇒1000=15n+n2
⇒n2+15n−1000=0
⇒n2+40n−25n−1000=0
⇒n(n+40)−25(n+40)=0⇒(n+40)(n−25)=0
∴ n=25 [∵n≠−40]
Hence, she will save Rs. 2000 in 25 months.