Math, asked by millan1830, 3 months ago

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

Answered by amitnrw
2

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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Answered by Anonymous
2

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:

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