3. The 5th term of an AP is 8 and the 10th term is -7, find the 20th term and 50th term of an AP.
I would like to know the 20th term and 50th term
To find the 20th term of an Arithmetic Progression (AP), we need to know the common difference (d) between consecutive terms in the sequence.
Given that the 5th term (a₅) is 8 and the 10th term (a₁₀) is -7, we can use these values to find the common difference.
Using the formula for the nth term of an AP:
aₙ = a₁ + (n-1)d
Let's substitute the known values to find the common difference:
8 = a₁ + (5-1)d
-7 = a₁ + (10-1)d
Simplify these equations:
8 = a₁ + 4d
-7 = a₁ + 9d
Now, we can solve these two equations simultaneously to find the values of a₁ (1st term) and d (common difference).
Subtract the first equation from the second equation:
(-7) - (8) = (a₁ + 9d) - (a₁ + 4d)
-7 - 8 = a₁ + 9d - a₁ - 4d
-15 = 5d
Divide by 5 on both sides:
d = -15/5
d = -3
Now substitute the value of d back into the equation:
8 = a₁ + 4(-3)
8 = a₁ - 12
a₁ = 20
Now that we know the first term (a₁ = 20) and the common difference (d = -3), we can find the 20th term (a₂₀) and the 50th term (a₅₀) of the AP using the formula:
aₙ = a₁ + (n-1)d
Substitute the known values:
a₂₀ = 20 + (20-1)(-3)
a₅₀ = 20 + (50-1)(-3)
Simplify these equations:
a₂₀ = 20 + 19(-3)
a₅₀ = 20 + 49(-3)
Calculate:
a₂₀ = 20 - 57
a₂₀ = -37
a₅₀ = 20 - 147
a₅₀ = -127
Therefore, the 20th term of the AP is -37 and the 50th term is -127.
To find the 20th term and 50th term of an arithmetic progression (AP), we need to use the given information about the 5th term and 10th term.
An arithmetic progression has a common difference, denoted by 'd', which is the fixed difference between any two consecutive terms in the sequence.
Let's start by finding the common difference:
We are given:
5th term (a5) = 8
10th term (a10) = -7
The formula to find the 'n-th' term of an AP is given by:
an = a1 + (n - 1) * d
Using this formula, we can find the common difference 'd':
a5 = a1 + (5 - 1) * d
8 = a1 + 4d ............ (equation 1)
a10 = a1 + (10 - 1) * d
-7 = a1 + 9d ............ (equation 2)
To eliminate 'a1', we'll multiply equation 1 by 9 and equation 2 by 4:
9 * 8 = 9a1 + 36d
-7 * 4 = 4a1 + 36d
72 = 9a1 + 36d ............ (equation 3)
-28 = 4a1 + 36d ............ (equation 4)
Subtracting equation 4 from equation 3:
72 - (-28) = (9a1 - 4a1) + (36d - 36d)
100 = 5a1
Dividing both sides by 5:
a1 = 20
Now that we have found the first term 'a1', we can substitute it back into equation 1 to find the common difference 'd':
8 = 20 + 4d
4d = -12
d = -3
Therefore, the first term (a1) is 20 and the common difference (d) is -3.
Now, we can use the formula an = a1 + (n - 1) * d to find the 20th and 50th terms:
a20 = 20 + (20 - 1) * (-3)
a20 = 20 + 19 * (-3)
a20 = 20 - 57
a20 = -37
Therefore, the 20th term of the arithmetic progression is -37.
a50 = 20 + (50 - 1) * (-3)
a50 = 20 + 49 * (-3)
a50 = 20 - 147
a50 = -127
Therefore, the 50th term of the arithmetic progression is -127.
probably best to first fins a and d:
a+4d = 8
a+9d = -7
now subtract, and crank it out.