Find the arithmetic sequence whose fifth term = 21, and the tenth term = 3 times the second term

by our definitions:

a + 4d = 21 ---> a = 21-4d **

a+ 9d = 3(a+d)
a + 9d = 3a + 3d
2a = 6d
a = 3d ***

equate ** and ***
3d = 21-4d
7d = 21

carry on

To find the arithmetic sequence, we first need to find the common difference. The common difference (d) is the constant difference between each term in the sequence.

Let's denote the first term as 'a' and the common difference as 'd'. We are given the following constraints:

Fifth term = 21
Tenth term = 3 times the second term

We can write the equations based on these constraints:

a + 4d = 21 (equation 1) -- Since the fifth term is 21, which means we have (a + 4d) as the fifth term.

a + 9d = 3(a + d) (equation 2) -- Since the tenth term is 3 times the second term, we have (a + 9d) as the tenth term and (a + d) as the second term.

We have a system of two equations with two variables. We can solve this system using the substitution method or the elimination method.

Let's use the substitution method:

From equation 1, we can express 'a' in terms of 'd':
a = 21 - 4d

Now, substitute this expression for 'a' in equation 2:

(21 - 4d) + 9d = 3((21 - 4d) + d)

Simplifying:

21 - 4d + 9d = 63 - 12d + 3d

Combine like terms:

21 + 5d = 63 - 9d

Move the variables to one side and the constants to the other side:

14d = 63 - 21

14d = 42

Divide both sides by 14:

d = 3

Now that we have the value of the common difference 'd', we can substitute it back into either equation 1 or equation 2 to find the value of 'a'. Let's substitute it into equation 1:

a + 4(3) = 21

a + 12 = 21

Subtract 12 from both sides:

a = 21 - 12

a = 9

Therefore, the first term of the arithmetic sequence is 9 and the common difference is 3. Hence, the arithmetic sequence is:

9, 12, 15, 18, 21, ...

Each term in the sequence is obtained by adding 3 to the previous term.