Kalpana saves some amount every month. In the first three months, she saves ₹ 100, ₹ 150, and ₹ 200 respectively. In how many months will she save ₹ 1200?

Activity :- Kalpana’s monthly saving is ₹ 100, ₹ 150, ₹ 200, ......, ₹ 1200

Here, d = 50. Therefore this sequence is an A.P.

a = 100, d = 50, t_{n} = `square`, n = ?

t_{n} = a + (n – 1) `square`

`square` = 100 + (n – 1) × 50

`square/50` = n – 1

n = `square`

Therefore, she saves ₹ 1200 in `square` months.

#### Solution

Kalpana’s monthly saving is ₹ 100, ₹ 150, ₹ 200, ......, ₹ 1200

Here, d = 50. Therefore this sequence is an A.P.

a = 100, d = 50, t_{n} = **1200**, n = ?

t_{n} = a + (n – 1) **d**

∴** 1200** = 100 + (n – 1) × 50

∴ 1200 – 100 = (n – 1) × 50

∴ 1100 = (n – 1) × 50

∴ **`1100/50`** = n – 1

∴ 22 = n – 1

n = **23**

Therefore, she saves ₹ 1200 in **23** months.