A high-interest savings account pays 5.5% interest compounded annually. If $300 is deposited initially and again at the first of each year, which summation represents the money in the account 10 years after the initial deposit?
Answers
Given : A high-interest savings account pays 5.5% interest compounded annually.
$300 is deposited initially and again at the first of each year,
To Find : money in the account 10 years after the initial deposit
Solution:
Money deposited every year = 300 $
P = 300
R = 5.5 %
A = P(1 + R/100)ⁿ
n = 10 for 1st deposit and n = 1 for last deposit
Money deposited in 1st year becomes = 300(1 + 5.5/100)¹⁰
Money deposited in 2nd year becomes = 300(1 + 5.5/100)⁹
and so on
Money deposited in 9th year becomes 300(1 + 5.5/100)²
Money deposited in 10th year becomes = 300(1 + 5.5/100)¹
Total Amount
= 300(1 + 5.5/100)¹ + 300(1 + 5.5/100)² + .. + .. + 300(1 + 5.5/100)⁹ + 300(1 + 5.5/100)¹⁰
This is an GP
where a = 300(1 + 5.5/100) = 300(1 + 55/1000) = 300(1.055)
r = 1.055
Sₙ = a(rⁿ - 1)/(r - 1)
= 300(1.055) ( 1.055¹⁰ - 1)/(1.055 - 1)
= 4,075
money in the account 10 years after the initial deposit = 4,075 $
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Answer:
Money deposited every year = 300 $
P = 300
R = 5.5 %
A = P(1 + R/100)ⁿ
n = 10 for 1st deposit and n = 1 for last deposit
Money deposited in 1st year becomes = 300(1 + 5.5/100)¹⁰
Money deposited in 2nd year becomes = 300(1 + 5.5/100)⁹
and so on
Money deposited in 9th year becomes 300(1 + 5.5/100)²
Money deposited in 10th year becomes = 300(1 + 5.5/100)¹
Total Amount
= 300(1 + 5.5/100)¹ + 300(1 + 5.5/100)² + .. + .. + 300(1 + 5.5/100)⁹ + 300(1 + 5.5/100)¹⁰
This is an GP
where a = 300(1 + 5.5/100) = 300(1 + 55/1000) = 300(1.055)
r = 1.055
Sₙ = a(rⁿ - 1)/(r - 1)
= 300(1.055) ( 1.055¹⁰ - 1)/(1.055 - 1)
= 4,075
money in the account 10 years after the initial deposit = 4,075 $
Step-by-step explanation: