Answers

2015-03-09T11:23:12+05:30
Sn=236000
f=2000
d=200
sn=
 \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
number of days=40
last instalment(xn)=dn+(f-d)
=200n+1800
=200*40+1800=9800 rs
2 5 2
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

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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

1 5 1