Question

A man repays a loan of rupees 3250 by paying rupees 20in the first month and then increases the payment by rupees 15 every month how long will it take him to clear a loan​

Answer 1

Answer:

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so here we go

Step-by-step explanation:

so basically in this sum

we have to use concept of arithmetic progression ie ap

so as the man increases payment by 15 rs each month

common difference between each of those years would be constant

so as common difference d is constant

it forms an ap

so here we have to find value of n ie number of years

so then a=t1=20 ie first term

d=20

sum of all payments ie Sn=3250

now use formula

Sn=n/2×(2a+(n-1)×d

substitute given values in it

we get 3250=n/2×((2×20)+(n-1)×15))

ie 6500=n×(40+15n-15)

=n×(25+15n)

6500=25n+15n square

so 15n square+25n-6500=0

divide through out by 5 we get

3n square+5n-1300=0

so this is a quadratic equation formed now

so comparing it with an square+an+c=0

we get a=3 b=5 c=-1300

so now use delta=b square-4ac

substitute a b and c values we get

delta=15625

so now using formula method

ie n=

$-b + - \sqrt{ b \: square - 4ac \div 2a}$

substitute values found till now we get

so we get by substituting given values

n=-5+125/2×3 or n=-5-125/2×3

ie n=120/6 or n=-130/6

ie n=20 or n=-65/2

but as no of years are never negative n=-65/2 is absurd

hence n=20

hence it will take him 20 years to clear a loan