The 4th term of an AP is 20 and the 8th term is 60. Find the 15th term
OR
an = a1 + ( n - 1 ) d
where
a1 = the initial term
d = the common difference of successive members
a4 = a1 + 3 d = 20
a8 = a1 + 7 d = 60
a8 - a4 = 60 - 20 = a1 + 7 d - ( a1 + 3 d )
40 = a1 + 7 d - a1 - 3 d
40 = a1 - a1 + 7 d - 3 d
40 = 4 d
d = 40 / 4
d = 10
Put this value in equation:
a1 + 3 d = 20
a1 + 3 ∙ 10 = 20
a1 + 30 = 20
Subtract 30 to both sides
a1 = - 10
a15 = a1 + 14 d
a15 = - 10 + 14 ∙ 10
a14 = - 10 + 140
a14 = 130
To find the 15th term of an arithmetic progression (AP), we need to determine the common difference (d) first.
The common difference (d) is the difference between any two consecutive terms of the AP.
Given that the 4th term (a₄) is 20 and the 8th term (a₈) is 60, we can use the formula for the nth term of an AP:
aₙ = a₁ + (n - 1)d
Let's find the common difference (d) first:
a₄ = a₁ + (4 - 1)d (substituting n = 4)
20 = a₁ + 3d
a₈ = a₁ + (8 - 1)d (substituting n = 8)
60 = a₁ + 7d
From these two equations, we can form a system of linear equations to solve for the values of a₁ and d.
Equation 1: 20 = a₁ + 3d
Equation 2: 60 = a₁ + 7d
By subtraction method, we can eliminate a₁:
(60 - 20) = (a₁ + 7d) - (a₁ + 3d)
40 = 4d
Simplifying the equation, we get:
d = 10
Now that we have the value of d, we can substitute it back into Equation 1 to find a₁:
20 = a₁ + 3(10)
20 = a₁ + 30
a₁ = 20 - 30
a₁ = -10
Now we have found the value of the first term (a₁) and the common difference (d).
To find the 15th term (a₁₅), we can use the formula:
aₙ = a₁ + (n - 1)d
Substituting the values we found:
a₁₅ = -10 + (15 - 1)(10)
a₁₅ = -10 + 14(10)
a₁₅ = -10 + 140
a₁₅ = 130
Therefore, the 15th term of the AP is 130.
To find the 15th term of an arithmetic progression (AP) when you are given the 4th and 8th terms, you need to first find the common difference of the AP.
The formula for the nth term (Tn) of an AP is given by:
Tn = a + (n - 1)d
Where:
a = first term of the AP
d = common difference of the AP
n = the term number
Given that the 4th term (T4) is 20 and the 8th term (T8) is 60, we can set up two equations:
T4 = a + (4 - 1)d = 20 -- (Equation 1)
T8 = a + (8 - 1)d = 60 -- (Equation 2)
Simplifying the equations, we get:
a + 3d = 20 -- (Equation 1)
a + 7d = 60 -- (Equation 2)
Now, we can solve these two simultaneous equations to find the values of a and d. Here's a step-by-step process to solve the equations using the method of substitution:
Step 1: Solve Equation 1 for a:
a = 20 - 3d
Step 2: Substitute the value of a in Equation 2:
(20 - 3d) + 7d = 60
Step 3: Simplify and solve for d:
20 + 4d = 60
4d = 60 - 20
4d = 40
d = 40/4
d = 10
Step 4: Substitute the value of d into Equation 1 to find a:
a + 3(10) = 20
a + 30 = 20
a = 20 - 30
a = -10
Thus, we have found that the first term (a) is -10 and the common difference (d) is 10.
Now, we can find the 15th term (T15) using the formula Tn = a + (n - 1)d:
T15 = -10 + (15 - 1) * 10
T15 = -10 + 14 * 10
T15 = -10 + 140
T15 = 130
Therefore, the 15th term of the arithmetic progression is 130.
4 differences between 4th and 8th terms
(60 - 20) / 4 = 10 ... this is the common difference
(15 - 8) * 10 = 70
15th term = 8th term + 70