Question

A man repays a loan of rs 3250 by paying rs20 in the first month and then increase the payment by rs15 every month. how long will it take him to clear the loan?

Answer 1

The question can be answered using Arithmetic Progression.

Here, S(n)= 3250
First Term (a)= 20
Common Difference (d) =15
No. of Terms (Months of payment) = ?

$s(n) = \frac{n}{2} \: (2a + (n - 1)d)$

$3250 = \frac{n}{2} (40 + (n - 1)15)$

On solving the above equation, we get,

$n = 20 \: \: \: n = - \frac{ 65}{3}$

Since, N cannot be negative.....
Hence, N=20. Man would repay the loan after 20 months.

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Answer 2

Answer:

Step-by-step explanation:

Solution :-

Here, a = 20,

d = 15 (As payment increases by Rs. 15)

and S(n) = 3

We know that,

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

Putting all the values, we get

⇒ 3250 = n/2[2 × 20 + (n - 1) (15)]

⇒ 3250 × 2 = n[40 + 15n - 15]

⇒ 6500 = n[25 + 15n]

⇒ 1300 = n[5 + 3n]

⇒ 3n² + 5n - 1300 = 0

⇒ 3n² + 65n - 60n - 1300 = 0

⇒ n(3n + 65) - 20(3n + 65) = 0

⇒ (n - 20) (3n + 65) = 0

⇒ n = 20, - 65/3 (As n can't be negative)

⇒ n = 20

Hence, men will take 20 months to clear the loans.

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