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?

#### Solution

Yasmeen, during the first month, saves = ₹ 32

During the second month, saves = ₹ 36

During the third month, saves = ₹ 40

Let Yasmeen saves Rs 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., S_{n} = ₹ 2000

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

`S_n = n/2[2a + (n - 1)d]`

⇒ `2000 = n/2 [2 xx 32 + (n - 1) xx 4]`

⇒ 2000 = n(32 + 2n – 2)

⇒ 2000 = n(30 + 2n)

⇒ 1000 = n (15 + n)

⇒ 1000 = 15n + n^{2}

⇒ n^{2} + 15n – 1000 = 0

⇒ n^{2} + 40n – 25n – 1000 = 0

⇒ n(n + 40) – 25(n + 40) = 0

⇒ (n + 40)(n – 25)= 0

∴ n = 25 .......[∵ n ≠ – 40]

Since, months cannot be negative

Hence, in 25 months she will save ₹ 2000.