Question

Raghu wish to buy a laptop . he can buy it by paying ₹ 40,000 cash or by giving it in 10 iñstallments as ₹ 4800 in the first month,₹4750 in the second month,₹ 4700 in the third month and so on . if he pays the money in this fashion, find 1) total amount paid in 10 ínstállments? 2) how much extra amount that he has to pay than the cost?

Answer 1

Given :

• Cost of Laptop = ₹ 40,000

• 10 Iñstallments

• ₹ 4800 - First month, ₹4750 - Second month, ₹ 4700 - Third month

To Find :

• Total amount paid in 10 Instállments

• The extra amount he pays in Instállments

Solution :

4800 + 4750 + 4700 + . . . . . + 10 terms

Here a = 4800

1) Total amount paid in 10 Instállments?

So, t₁ = 4800, t₂ = 4750

d = t₂ - t₁

d = 4750 - 4800

d = - 50

And, n = 10

By using formula,

⇒ $Sn = \dfrac{n}{2} (2a + (n - 1)d)$

⇒ $S10 = \dfrac{10}{2} (2 \times 4800 + (10 - 1) - 50)$

⇒ $S10 = \dfrac{10}{2} (2 \times 4800 + 9x - 50 )$

⇒ $S10 = 5 (2 \times 4800 + 9x - 50 )$

⇒ $S10 = 5 (9600 - 450 )$

⇒ $S10 = 5 × 9150$

⇒ $S10 = 45750$

•°• Total amount paid in 10 Instállments = ₹45750

2) The extra amount he pays in Instállments :

To find the extra amount, we should subtract the cost of laptop from total amount paid in 10 Instállments

⇒Total amount paid in 10 Instállments - Cost of Laptop

⇒ ₹45750 - ₹40,000

⇒₹5750

•°• The extra amount he pays in Instállments = ₹5750