find the 15th term of a linear sequence, whose 5th term is 30 and 20th term is 75.

Recall the formular

Tn=a+(n-1)d
n=15
T15=a+(15-1)d
=a+14d
but 5th term=30
30=a+(5-1)d
30=a+4d ........(i)
20th term=75
75=a+(20-1)d
75=a+19d........(ii)
from (i) make a the subject of the formular
30=a+4d
a=30-4d.........(iii)
sub (iii) into (ii)
75=30-4d+19d
C.L.T
75-30=-4d+19d
45/15=15d/15
d=3
sub d into (i)
30=a+4*3
30=a+12
a=30-12
a=18
therefore T15=18+14*3
=60

T15 is 2/3 of the way from T5 to T20

So, its value will be 2/3 of the way from 30 to 75.

Or, note that
T20 = T5 + 15d
So, 15d = 75-30 = 45
d = 15

T15 = T5 + 10d

To find the 15th term of a linear sequence, we can use the formula for the nth term of a linear sequence, which is:

a(n) = a(1) + (n - 1)d

where a(n) is the nth term, a(1) is the first term, n is the position of the term, and d is the common difference between terms.

Given that the 5th term is 30, we can substitute this information into the formula:

30 = a(1) + (5 - 1)d

Simplifying this equation, we get:

30 = a(1) + 4d ----- (equation 1)

Similarly, we can use the information that the 20th term is 75:

75 = a(1) + (20 - 1)d

Simplifying this equation, we get:

75 = a(1) + 19d ----- (equation 2)

We now have a system of two equations with two unknowns (a(1) and d).

To solve this system, we can subtract equation 1 from equation 2:

75 - 30 = (a(1) + 19d) - (a(1) + 4d)

45 = 15d

Dividing both sides of the equation by 15, we get:

d = 3

We can now substitute this value of d into equation 1 to find a(1):

30 = a(1) + 4(3)

30 = a(1) + 12

Subtracting 12 from both sides of the equation, we get:

18 = a(1)

Now that we have found the value of a(1) as 18 and the common difference d as 3, we can use the formula for the nth term to find the 15th term:

a(15) = 18 + (15 - 1)(3)

Simplifying this equation, we get:

a(15) = 18 + 14(3)

a(15) = 18 + 42

a(15) = 60

Therefore, the 15th term of the linear sequence is 60.

To find the 15th term of a linear sequence, we need to find the common difference, which is the difference between consecutive terms in the sequence. Once we have the common difference, we can find any term in the sequence using the formula:

Term(n) = First term + (n - 1) * Common difference

Let's find the common difference first:

Given that the 5th term is 30, we can write the equation as:

Term(5) = First term + (5 - 1) * Common difference
30 = First term + 4 * Common difference

Similarly, for the 20th term:

Term(20) = First term + (20 - 1) * Common difference
75 = First term + 19 * Common difference

We now have a system of two equations:

First term + 4 * Common difference = 30
First term + 19 * Common difference = 75

Let's solve this system of equations:

First, we subtract the first equation from the second equation:

(First term + 19 * Common difference) - (First term + 4 * Common difference) = 75 - 30
15 * Common difference = 45
Common difference = 45 / 15
Common difference = 3

Now that we know the common difference is 3, we can find the first term by substituting this value back into one of the original equations. Let's use the first equation:

First term + 4 * Common difference = 30
First term + 4 * 3 = 30
First term + 12 = 30
First term = 30 - 12
First term = 18

Finally, we can find the 15th term using the formula:

Term(15) = First term + (15 - 1) * Common difference
Term(15) = 18 + 14 * 3
Term(15) = 18 + 42
Term(15) = 60

Therefore, the 15th term of the linear sequence is 60.