If a certain sum, invested under compound interest, amounts to twice as much at the end of the 7th year as it would at the end of the 2nd year, then the amount at the end of the 64th year will be how many times that at the end of the 49th year.
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Amount at the end of the 64th year will be 4√2 times ( 5.657 times) that at the end of the 49th year. if Amount at end of 7th year is double of at end of 2nd year
Step-by-step explanation:
Let say Amount = A
Amount at end of 2nd Year = A (1 + R/100)
Amount at and of 7th year = A(1 + R/100)⁷
A(1 + R/100)⁷ = 2A (1 + R/100)
=> (1 + R/100)⁶ = 2
amount at the end of the 64th year = A(1 + R/100)⁶⁴
amount at the end of the 49th year = A(1 + R/100)⁴⁹
A(1 + R/100)⁶⁴ / A(1 + R/100)⁴⁹
= (1 + R/100)¹⁵
= ( 1 + R/100)¹² * (1 + R/100)³
= (( 1 + R/100)⁶)² * √((1 + R/100)⁶)
= 2² * √2
= 4√2 times
= 5.657 times
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