To solve the question \( A = P \left( 1 + \frac{...}{100} \right)^n \), we need to identify what the blank space represents in the context of compound interest.
### Step-by-Step Solution:
1. **Understand the formula**: We start with the formula for compound interest, which is given by:
\[
A = P \left( 1 + \frac{R}{100} \right)^n
\]
where:
- \( A \) is the amount after time \( n \),
- \( P \) is the principal amount (initial investment),
- \( R \) is the rate of interest per annum,
- \( n \) is the number of years the money is invested or borrowed.
2. **Identify the components**: In the given formula \( A = P \left( 1 + \frac{...}{100} \right)^n \):
- The blank space represents the rate of interest \( R \).
3. **Fill in the blank**: Therefore, we can fill in the blank with \( R \):
\[
A = P \left( 1 + \frac{R}{100} \right)^n
\]
4. **Conclusion**: The complete formula with the filled blank is:
\[
A = P \left( 1 + \frac{R}{100} \right)^n
\]
### Final Answer:
The blank should be filled with \( R \) (the rate of interest per annum).
---
Topper's Solved these Questions
COMPOUND INTEREST
RS AGGARWAL|Exercise TEST PAPER-11|10 Videos
AREA OF A TRAPEZIUM AND A POLYGON
RS AGGARWAL|Exercise TEST PAPER - 18 C (Fill in the blanks)|1 Videos
CONSTRUCTION OF QUADRILATERALS
RS AGGARWAL|Exercise Test Paper-17 (E)|1 Videos
Similar Questions
Explore conceptually related problems
Formula 1 Let P be the principal and the rate of interest be R% per annum.If the interest is compunded anually then the amount A and the compound interest C.I.at the end of n years are given by nA=P(1+(R)/(100))^(n) and C.I.=A-P=P{(1+(R)/(100))^(n)-1} respectively.
The compound interest on a certain sum is given by C.I.=P(1+(R)/(100))^(n)-P . Find C.I. when P=rs1000,R=10% P.a., and n=2. The following steps are involved in solving the above problem. Arrange them in sequential order. (A) therefore C.I.=rs210 (B) 1000((11)/(10))((11)/(10))-1000=1210-100 (C) Given Cl=P(1+(R)/(100))^(n)-P,P=rs1000 , R=10% p.a., and n=2 (D) C.I. = 1000+(1+(10)/(100))^(2)-1000
Formula 2 Let P be the principal and the rate of interest be R% per annum.If the interest is compounded annually then the amount A and the compound interest C.I.at the end of n years are given by A=P(1+(R)/(100k))^(nk) and C.I.A-P=P{((1_(R))/(100k))^(nk)-1} respectively
Formula 1 Let P be the population of a city or a town at the beginning of a certain year and the population grows at a constant rate of R% per annum then population after n years =P(1+(R)/(100))^(n)
Statement I: If the annual increase in the population be 20% and present population be 10,000 then the population after 4 years is 20,736 .Statement II: the original population of a location be p and the annual growth rate be r%. The population after n years is P(1+(r)/(100))^(n).
Statement-1: The number of zeros at the end of 100! Is, 24. Statement-2: The exponent of prine p in n!, is [(n)/(p)]+[(n)/(p^(2))]+.......+[(n)/(p^(r))] Where r is a natural number such that P^(r)lenltP^(r+1) .
prove that 1P_(1)+2.2P_(2)+3.3P_(3)+.......+n.nP_(n)=(n+1)P_(n+1)-1
For all p, such that 1<=p<=100 if n(A_(p))=p+2 and A_(1)sub A_(2)sub A_(3)sub......sub A_(100) and nnn_(p=1)^(100)A_(p)=A then n(A)=
RS AGGARWAL-COMPOUND INTEREST -TEST PAPER-11 ( Fill in the blanks )