If $1,400 is invested in a savings account offering interest at a rate of 4.5% per year, compounded continuously, how fast is the balance growing after 4 years? $___ per year
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Compounding continuously means interest for each day is compounded every day. So interest is compounded 365 times a day. I am ignoring the effect of one day extra in the leap year that will come in the 4 fours.
Interest per day = r = 4.5 /(365 * 100 ) = $ 0.0001233 /day
Number of times the interest is compounded = n = 365 * 4 = 1460
Accumulated total after 4 years = S = P (1 + r )¹⁴⁶°
= $ 1,400 (1 + 0.0001233)¹⁴⁶° = $ 1, 676.08
Continuous compounding is also obtained by the formula =
= P e^nr, where e = 2.718
= $ 1400 * 2.718^(4.5*4/100) = $ 1,676.10 (almost same as above)
Growth per year at the end of 4th year = S_4 * 4.5/100 = $ 75.42 / year
Interest per day = r = 4.5 /(365 * 100 ) = $ 0.0001233 /day
Number of times the interest is compounded = n = 365 * 4 = 1460
Accumulated total after 4 years = S = P (1 + r )¹⁴⁶°
= $ 1,400 (1 + 0.0001233)¹⁴⁶° = $ 1, 676.08
Continuous compounding is also obtained by the formula =
= P e^nr, where e = 2.718
= $ 1400 * 2.718^(4.5*4/100) = $ 1,676.10 (almost same as above)
Growth per year at the end of 4th year = S_4 * 4.5/100 = $ 75.42 / year
kvnmurty:
thanx n u r welcom
S_4 typing mistake. It is S = $1,676.10
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