# 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

2
by deeeeep

2015-03-09T11:23:12+05:30
Sn=236000
f=2000
d=200
sn=

number of days=40
last instalment(xn)=dn+(f-d)
=200n+1800
=200*40+1800=9800 rs
what f here stands for ?
first term..first install ment
any dout..comment here
leave dat pls ans dis one
Cards marked with numbers 5 to 100 are placed.one card is drawn from this . find the probblity that the no. on the card is
1) a perfect square
2) a multiple of 7
3 ) a prime no. less than 25
2015-03-09T13:00:26+05:30

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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