Answer:
[tex]T_{2020} = 15[/tex]
Explanation:
Given
[tex]T_1 = 10[/tex]
[tex]T_2 = 20[/tex]
Each term after the second term is the average of all of the preceding terms
Required:
Explain how to solve the 2020th term
Solve the 2020th term
Solving the 2020th term of a sequence using conventional method may be a little bit difficult but in questions like this, it's not.
The very first thing to do is to solve for the third term;
The value of the third term is the value of every other term after the second term of the sequence; So, what I'll do is that I'll assign the value of the third term to the 2020th term
This is proved as follows;
From the question, we have that "..... each term after the second term is the average of all of the preceding terms", in other words the MEAN
[tex]T_{n} = \frac{\sum T{k}}{n-1} ; where: k = 1 .... n -1[/tex]
Assume n = 3
[tex]T_{3} = \frac{T_1 + T_2}{2}[/tex]
Multiply both sides by 2
[tex]2 * T_{3} = \frac{T_1 + T_2}{2} * 2[/tex]
[tex]2T_{3} = T_1 + T_2[/tex]
Assume n = 4
[tex]T_{4} = \frac{T_1 + T_2 + T_3}{3}[/tex]
[tex]T_{4} = \frac{(T_1 + T_2) + T_3}{3}[/tex]
Substitute [tex]2T_{3} = T_1 + T_2[/tex]
[tex]T_{4} = \frac{2T_3 + T_3}{3}[/tex]
[tex]T_{4} = \frac{3T_3}{3}[/tex]
[tex]T_{4} = T_3[/tex]
Assume n = 5
[tex]T_{5} = \frac{T_1 + T_2 + T_3 +T_4}{4}[/tex]
[tex]T_{5} = \frac{(T_1 + T_2) + T_3 +(T_4)}{4}[/tex]
Substitute [tex]2T_{3} = T_1 + T_2[/tex] and [tex]T_{4} = T_3[/tex]
[tex]T_{5} = \frac{2T_3 + T_3 +T_3}{4}[/tex]
[tex]T_{5} = \frac{4T_3}{4}[/tex]
[tex]T_{5} = \frac{(5-1)T_3}{5-1}[/tex]
Replace 5 with n
[tex]T_{n} = \frac{(n-1)T_3}{n-1}[/tex]
(n-1) will definitely cancel out (n-1); So, we're left with
[tex]T_{n} = T_3[/tex]
Hence,
[tex]T_{2020} = T_3[/tex]
Calculating [tex]T_3[/tex]
[tex]T_{3} = \frac{10 + 20}{2}[/tex]
[tex]T_{3} = \frac{30}{2}[/tex]
[tex]T_{3} = 15[/tex]
Recall that [tex]T_{2020} = T_3[/tex]
[tex]T_{2020} = 15[/tex]
