Question

i want the ans fast of this question

Answer 1

Answer:

We assume the the man clears the loan in nth month. We find the total amount paid up to nth month as using sum up to n terms of an AP Sn=n2{2a+(n−1)d} where a the amount is is paid in first month and d is the increase amount every month and equate to given 3250. We solve for n to get the required number of months.  

Complete step by step answer:

We know that the sum up to nth of an arithmetic progression with first term a and common difference d is given by

Sn=n2{2a+(n−1)d}

We are given in the question that the man repays a loan of Rs 3250 by paying Rs 20 in the first month and then increases the payment by Rs 15 every month. So in second month he pays 20+15=35 rupees and in the third month he pays 35+15=50 rupees and so on. So we get sequence of payments as

20,35,50,...

We see that the above sequence is an AP sequence with common difference d=35−20=50−35=15 and first term a=20 . Let us assume that he clears the loan in nth month which means he pays back complete 3250 rupees. So the sum of the amounts from the first month to nth month will be equal to 3250 rupees. We use the sum up to nth term formula for an AP and equate the sum to 3250. We have;

3250=n2{2×20+(n−1)15}⇒6500=n(15n+25)⇒15n2+25n−6500=0

We divide both sides by 5 to have;

⇒3n2+5n−1300=0

We see that the above equation is quadratic equation in n which we solve by splitting the middle term 5n as;

⇒3n2−60n+65n−1300=0⇒3n(n−20)+65(n−20)=0⇒(n−20)(3n+65)=0⇒n=20 or n=−653

Since the number of months cannot be negative we reject the solution n=−653 and accept the solution as n=20 . So the man will take 20 months to clear the loan. Hence the correct choice is C.

Note:

We note that arithmetic progression; abbreviate d as AP is a type of sequence where the difference between any two consecutive numbers is constant. We can also find the amount he paid in the last month using formula for nth of an AP as tn=a+(n−1)d .We also note that the question is silent about interest and hence we should not use simple interest formula. We can split the middle term in the quadratic equation from prime factorization of 3×1300 . We can also use quadratic formula but we need to find the square root there,

Step-by-step explanation: