The 10th term and 15th of an arithmetic progression is -5th and -71/2. Find the sum of the 20th term
Well, that sounds like quite the arithmetic progression! Let's break it down, shall we?
We are given that the 10th term is -5 and the 15th term is -71/2.
To find the common difference, we can subtract the 10th term from the 15th term.
Let's do some math: (-71/2) - (-5)
When we subtract a negative number, it's like adding the positive version of that number. So, (-71/2) - (-5) is the same as (-71/2) + 5.
Now, let's simplify: (-71/2) + 5 = -71/2 + 10/2 = (-71 + 10)/2 = -61/2
So, the common difference is -61/2.
To find the 20th term, we can use the formula: term = first term + (n-1) * common difference, where n is the term number.
Let's plug in the values: term = -5 + (20-1) * (-61/2)
Now, let's do some more math: term = -5 + 19 * (-61/2)
To make it a little easier, let's simplify first: term = -5 + (-1159/2)
Now, let's add those fractions: term = -10/2 + (-1159/2) = -1169/2
So, the 20th term is -1169/2.
Now, to find the sum of the 20th term, we can use the formula for the sum of an arithmetic progression: sum = (n/2) * (first term + last term), where n is the number of terms.
Let's plug in the values: sum = (20/2) * (-5 + (-1169/2))
Let's simplify some more: sum = 10 * (-5 - (1169/2)) = 10 * (-10 - 1169)/2 = 10 * (-1179)/2
Finally, let's multiply that out: sum = -11790/2 = -5895
So, the sum of the 20th term is -5895.
I hope that puts a smile on your face, even if arithmetic progressions aren't always the most exciting topic!
To find the sum of the 20th term of an arithmetic progression, we need to first determine the common difference (d) of the progression.
Given:
10th term (a₁₀) = -5th
15th term (a₁₅) = -71/2
Step 1: Finding the common difference (d)
We know that the formula for the nth term (aₙ) of an arithmetic progression is given by:
aₙ = a₁ + (n - 1)d
Let's use this formula to find the common difference:
a₁₀ = a₁ + (10 - 1)d (substituting n = 10)
-5 = a₁ + 9d
a₁₅ = a₁ + (15 - 1)d (substituting n = 15)
-71/2 = a₁ + 14d
We now have a system of equations:
a₁ + 9d = -5 ...(Equation 1)
a₁ + 14d = -71/2 ...(Equation 2)
Step 2: Solving the system of equations
Subtracting Equation 1 from Equation 2, we get:
14d - 9d = -71/2 - (-5)
5d = -71/2 + 10/2
5d = -61/2
d = (-61/2) / 5
d = -61/10
d = -6.1
Step 3: Finding the 20th term (a₂₀)
Using the formula for the nth term of an arithmetic progression:
a₂₀ = a₁ + (20 - 1)d
a₂₀ = a₁ + 19d
Substituting the value of d:
a₂₀ = a₁ + 19(-6.1)
a₂₀ = a₁ - 115.9
Step 4: Finding the sum of the 20th term
The sum of n terms in an arithmetic progression is given by:
Sn = n/2 * (2a₁ + (n - 1)d)
Substituting the values:
S₂₀ = 20/2 * (2a₁ + (20 - 1)(-6.1))
S₂₀ = 10 * (2a₁ - 19(6.1))
S₂₀ = 10 * (2a₁ - 115.9)
To find the value of S₂₀, we need the value of a₁. Unfortunately, the given information does not provide the value of a₁.
To find the sum of the 20th term of an arithmetic progression, we need to determine the formula for the nth term of the sequence.
The formula for the nth term of an arithmetic progression is given by:
an = a1 + (n-1)d
Where:
an = nth term
a1 = first term
d = common difference between consecutive terms
Let's analyze the given information:
1) The 10th term, a10, is -5
2) The 15th term, a15, is -7/2
Using the formula, let's substitute these values:
a10 = a1 + 9d = -5
a15 = a1 + 14d = -7/2
We now have two equations:
a1 + 9d = -5 --(1)
a1 + 14d = -7/2 --(2)
Next, we'll solve these two equations to find the values of a1 and d.
From equation (1), we can find a1 in terms of d:
a1 = -5 - 9d
Substituting this in equation (2):
(-5 - 9d) + 14d = -7/2
Simplifying the equation:
-5 + 5d = -7/2
5d = -7/2 + 10/2
5d = 3/2
d = 3/10
Now, we have the value of the common difference, d. Let's find the first term, a1:
a1 = -5 - 9d
a1 = -5 - 9(3/10)
a1 = -5 - 27/10
a1 = -50/10 - 27/10
a1 = -77/10
Therefore, the first term, a1, is -77/10 and the common difference, d, is 3/10.
Now, to find the sum of the 20th term, we use the formula for the sum of an arithmetic progression:
Sn = n/2 [2a1 + (n-1)d]
Substituting the values we found:
n = 20
a1 = -77/10
d = 3/10
Sn = 20/2 [2(-77/10) + (20-1)(3/10)]
= 10 [-77/5 + 19(3/10)]
= 10 [-77/5 + 57/5]
= 10 [-20/5]
= 10 (-4)
= -40
Therefore, the sum of the 20th term is -40.