Question

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.

Answer 1

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 = \[\boxed{1200}\], n = ?

t_n = a + (n – 1) \[\boxed{\text{d}}\]

∴ \[\boxed{1200}\] = 100 + (n – 1) × 50

∴ 1200 – 100 = (n – 1) × 50

∴ 1100 = (n – 1) × 50

∴ \[\frac{\boxed{1100}}{50}\] = n – 1

∴ 22 = n – 1

n = \[\boxed{23}\]

Therefore, she saves ₹ 1200 in \[\boxed{23}\] months.