Question

sahil repays the total loan of rupees 2,36,000 by paying every month starting from the first instalment of rupees 2000. he increases the instalment rupees 200 every month. what amount would he pay as the last instalment of loan

Answer 1

sn=236000
f=2000
d=200
sn=
[tex] \frac{n}{2} (f+xn)=236000 \\ \frac{n}{2} (2000+dn+(f-d)]=236000 \\ n[2000+200n+1800]=236000*2 \\ n(3800+200n)=472000 \\ 200n ^{2} +3800n=472000 \\ 2 n^{2}+38n=4720 \\ n^{2} +19n=2360 \\ n^{2} +19n+90.25=2360+90.25=2450.25 \\ \\ (n+9.5) ^{2} =2450.25 \\ n+9.5=49.5 \\ n=40[/tex]
number of days=40
last instalment(xn)=dn+(f-d)
=200n+1800
=200*40+1800=9800 rs

Answer 2

Loan = P = Rs 2,36,000
First Installment of repayments = a = Rs 2,000
Increment of loan repayments = d = Rs 200
   Here we assume that there is no interest as per simple interest or compound interest.

  Sum = 2, 36, 000 = 2000 + 2200 + 2400 + ....
                         = a + (a + d) + (a + 2d) + ..  n terms  +  last installment
     Sum of AP series = [ 2a + (n-1) d] n/2 = [ 4000 + 200(n-1) ] n/2

we have to find the maximum value of n such that
           1900 n + 100 n² < 2,36,000
             n² + 19 n - 2360  < 0
factorizing:
        2360 = 59 * 4 * 10        =>  59 - 40 = 19
           (n + 59 ) (n - 40) <  0
           so    -59 < n < 40

We take the value of n as 39.
          So sum of AP series : Rs [4, 000 + 200 * 38 ] * 39/2
                  = Rs 5, 800 * 39 = Rs 2,26,200

The last installment = Rs 2,36,000 - Rs 2,26,200 = Rs Rs 9,800