Question

A man repay a loan of ₹ 3250 lay paying ₹20 in the first month and then increases the payment ₹15 ever month how loan will it take him to clear the loan

Answer 1

1st month lay p.20
remaining =3230
3230/35=92.28
92.28+1 month

Answer 2

Answer: n = 20

Step-by-step explanation:

a = 20, d = 15

Sn = 3250

$\bf\huge S_{n} = \frac{n}{2} [2a + (n - 1)d]$

$\bf\huge 3250 = \frac{n}{2} [40 + (n - 1)\times 15]$

⇒ 3250 × 2 = n[40 + 15n - 15]

⇒ 6500 = n[25 + 15n]

⇒ 1300 = n[5 + 3n]

⇒ 3n^2 + 65n - 60n - 1300 = 0

⇒ n(3n + 65) - 20(3n + 65) = 0

⇒ (n - 20)(3n + 65) = 0

Either n = - 65/3, (not possible)   or n = 20

Man will repay loan after 20 months