The first few multiples of 6 are [6, 12, 18, 24, 30, 36, 42, 48, 54, 60] and the first few multiples of 8 are [8, 16, 24, 32, 40, 48, 56, 64, 72, 80]. The first few multiples they share are [24, 48, 72, 96] making 24 the smallest multiple 6 and 8 have in common.

If the tiger has consumed 36 pounds of food in 6 days that's \( \frac{36}{6} \) = 6 pounds of food per day. The tiger needs to consume 54 - 36 = 18 more pounds of food to reach 54 pounds total. At 6 pounds of food per day that's \( \frac{18}{6} \) = 3 more days.

To add these fractions, first find the lowest common multiple of their denominators. The first few multiples of 3 are [3, 6, 9, 12, 15, 18, 21, 24, 27, 30] and the first few multiples of 11 are [11, 22, 33, 44, 55, 66, 77, 88, 99]. The first few multiples they share are [33, 66, 99] making 33 the smallest multiple 3 and 11 share.

Next, convert the fractions so each denominator equals the lowest common multiple:

\( \frac{6 x 11}{3 x 11} \) + \( \frac{6 x 3}{11 x 3} \)

\( \frac{66}{33} \) + \( \frac{18}{33} \)

Now, because the fractions share a common denominator, you can add them:

\( \frac{66 + 18}{33} \) = \( \frac{84}{33} \) = 2\(\frac{6}{11}\)

The equation for this sequence is:

a_n = a_n-1 + 4

where n is the term's order in the sequence, a_n is the value of the term, and a_n-1 is the value of the term before a_n. This makes the next number:

a₆ = a₅ + 4
a₆ = 17 + 4
a₆ = 21

A rational number (or fraction) is represented as a ratio between two integers, a and b, and has the form \({a \over b}\) where a is the numerator and b is the denominator. An improper fraction (\({5 \over 3} \)) has a numerator with a greater absolute value than the denominator and can be converted into a mixed number (\(1 {2 \over 3} \)) which has a whole number part and a fractional part.