Advertisements
Advertisements
प्रश्न
What will Rs.500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?
Advertisements
उत्तर
P = ₹ 500
C.I. = 10%
Amount after 1 year = `"P"(1 + "r"/100)`
= `500 + (1 + 10/100)`
= `500(11/10)`
Amount after 2 years = `500(11/10)^2`
Now a = `500 (11/10)`,
r = `(500(11/10)^2)/(500(11/10))`
= `11/10`
tn = arn–1
t10 = `500(11/10)(11/10)^9`
= `500(11/10)^10`
= 500(1.1)10
= 500(2.5937)
= 1296.87
Aliter:
Amount after 1 year = `500(1 + 10/100) = 500(11/10)`
Amount after 2 years = `500(11/10)(11/10) = 500(11/10)^2`
So amount after 10 years = `500(11/10)^10` = 1296.87
APPEARS IN
संबंधित प्रश्न
Find the sum of the first 20-terms of the arithmetic progression having the sum of first 10 terms as 52 and the sum of the first 15 terms as 77
Find the sum up to the 17th term of the series `1^3/1 + (1^3 + 2^3)/(1 + 3) + (1^3 + 2^3 + 3^3)/(1 + 3 + 5) + ...`
Compute the sum of first n terms of the following series:
8 + 88 + 888 + 8888 + ...
Compute the sum of first n terms of the following series:
6 + 66 + 666 + 6666 + ...
Find the general term and sum to n terms of the sequence `1, 4/3, 7/9, 10/27, ......`
Show that the sum of (m + n)th and (m − n)th term of an AP. is equal to twice the mth term
A man repays an amount of Rs.3250 by paying Rs.20 in the first month and then increases the payment by Rs.15 per month. How long will it take him to clear the amount?
In a certain town, a viral disease caused severe health hazards upon its people disturbing their normal life. It was found that on each day, the virus which caused the disease spread in Geometric Progression. The amount of infectious virus particle gets doubled each day, being 5 particles on the first day. Find the day when the infectious virus particles just grow over 1,50,000 units?
Choose the correct alternative:
If nC10 > nCr for all possible r, then a value of n is
Choose the correct alternative:
The sum up to n terms of the series `sqrt(2) + sqrt(8) + sqrt(18) + sqrt(32) + ...` is
