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 = ${\displaystyle \square }$, n = ?

t_n = a + (n – 1) ${\displaystyle \square }$

${\displaystyle \square }$ = 100 + (n – 1) × 50

${\displaystyle \frac{\square }{50}}$ = n – 1

n = ${\displaystyle \square }$

Therefore, she saves ₹ 1200 in ${\displaystyle \square }$ 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 = 1200, n = ?

t_n = a + (n – 1) d

∴ 1200 = 100 + (n – 1) × 50

∴ 1200 – 100 = (n – 1) × 50

∴ 1100 = (n – 1) × 50

∴ ${\displaystyle \frac{1100}{50}}$ = n – 1

∴ 22 = n – 1

n = 23

Therefore, she saves ₹ 1200 in 23 months.