1) Find the term of 12, 6, 3, 3/2, ...

2) find the 11th term if the 2nd is 8 and the 5th is 64.

3) if one-fourth of the air in a tank is removed by each stroke of an air pump, find the fractional part of the air remaining after seven strokes of the pump

4) a cask containing 20 liters of wine was emptied of one fifth of its contents and then filled with water. If this is done six times, how many liters of wine remain in the cask?

 

1. Formula for n^th term in geometric sequence a_n = ar^n-1 a₁₀ = 3/128
2. If arithmetic sequence, a11 = 176 ; If geometric sequence, a11 = 4096
3. 2187/16384 or in decimal form 0.1335 of the total air is removed in 7 strokes.
4. 0.23058 Liters

Step-by-step explanation

1. You forgot to put the nth term that needed to find. Just in case I put the formula of the geometric sequence. r = t?/t? ; r = 6/12 = 1/2 ; a_n = ar^n-1 Where a is first term and r is common ratio. Sample to find the 10th Term substitute n = 10, a = 12, r = 1/2 in a_n. a₁₀ = 12(1/2)^10-1= 12(1/2⁹)= 12/512= 3/128 ; a₁₀ = 3/128.
2. Step-by-step explanation: Please specify what sequence next time.

Assuming that it's an arithmetic sequence,

a2 = a1+d = 8

a5 = a1+4d = 64

a2 - a5 = 3d ; 3d = 56 ; d = 56/3 ; a1 = -32/3

a11 = a1 + 10d ; a11 = -32/3 + 10(56/3) ; a11 = 176

Assuming that it's a geometric sequence,

a2 = a1(r) = 8

a5 = a1(r)^4 = 64 ;

a5/a2 = r^3 ; r^3 = 64/8 ;r^3 = 8 ; r = 2 ; a1 = 4

a11 = a1(r)^10 ; a11 = 4(2)^10 ; a11 = 4096

3.I think it is simpler to calculate how much is left after each pump. After the first pump, 1/4 of the air is removed leaving 3/4, 1/4 of that is 3/16 leaving 9/16, 1/4 of that is 9/64 leaving 27/64, etc. You should be able to see that you are getting powers of 3/4. After n pumps, ?(3/4)n? of the air is left. ?(1−1/4)n=(3/4)n? = ?(3/4)7? = 2187/16384 or in decimal 0.1335.

4. (1-1/5) = 4/5 or 0.8 after each drain-fill cycle, there is 0.8 the amount of wine as before.

So, after 20 cycles, the wine remaining is 20(0.8)²⁰= 0.23058 Liters

ALTERNATIVE SOLUTION;

First term=20-1/5(20)=16

Second term= 12.8

r=4/5

A_n=A₁ (rⁿ-1)

A₂₀=16(4/5)^20-1

A₂₀=0.23058 Liters