a. geometricb. g[h] = (3/4)Ā·g[h-1]; g[1] = 384c. g[h] = 384Ā·(3/4)^(h-1)d. g[12] ā 16.2183 gallonse. g[12] ā 64.8732 quarts
step-by-step explanation:
part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.
part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.
g[h] = (3/4)Ā·g[h-1]
g[1] = 384 . . . . initial value required for a recursive definition
part c. the general formula for a geometric sequence is
a[n] = a[1]Ā·r^(n-1)
we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is
g[h] = 384Ā·(3/4)^(h-1)
part d. filling in the value and doing the arithmetic, we have
g[12] = 384Ā·(3/4)^11 ā 384Ā·0.042235136 ā 16.2183 . . gallons
part e. each gallon is 4 quarts, so
16.2183 gal Ć 4 qt/gal = 64.8732 qt
