Question

Mr. Jamison deposited $100 into a new savings account on January 1. On the first day of each month thereafter, he deposited three times the amount he deposited in the previous month. On June 15 of the same year, the total amount Mr. Jamison has deposited is $

Answer 1

Answer:

The total amount Mr. Jamison has deposited is $36400.

Step-by-step explanation:

Amount deposited by Mr. Jamison on January 1 = 100

Amount deposited on February 1 = 3(100)

Amount deposited on March 1 = 3[3(100)] = $3^{2} (100)$

Amount deposited on April 1 = $3[3^{2} (100)]=3^{3} (100)$

Amount deposited on May 1 = $3[3^{3} (100)]=3^{4} (100)$

Amount deposited on June 1 = $3[3^{4} (100)]=3^{5} (100)$

So, the total amount deposited by him on June 15 = $100+3(100)+3^{2}(100)+3^{3}(100)+3^{4}(100)+3^{5}(100)$

$= 100(1 + 3 + 3^{2}+3^{3}+3^{4}+3^{5} )$

The expression inside the parantheses is in Geometric Progression.

So,

$Amount deposited = 100(\frac{3^{6}-1 }{3-1} )$

= 50(728)

= 36400

Hence, the total amount deposited by Mr. Jamison on June 15 = $36,400.

Answer 2

Answer:

$36,400

Step-by-step explanation: