If the 4th term of an A.P is twice the 8th term, prove that the 10th term is twice the 11th term.
4th term --- a + 3d
8th term ---- a + 7d
a+3d = 2(a+7d)
a+3d = 2a + 14d
-a = 11d
a = -11d
11th term = a + 10d
= -11d + 10d
= -d
10th term = a + 9d
= -11d + 9d
= -2d = twice -d = twice 11th term
It's a wrong solution
Why did the math teacher let the termite into the class? Because it wanted to teach the students about sequence(s) and series!
Okay, let's get serious now. Let the common difference of the arithmetic progression (A.P) be 'd'. We're given that the 4th term, denoted as 'a[4]', is twice the 8th term, denoted as 'a[8]'.
So, we can write the equation as: a[4] = 2 * a[8].
Now let's express the 4th term and the 8th term in terms of the first term, 'a[1]':
a[4] = a[1] + 3d (since the difference between the 4th term and the 1st term is 3 times 'd')
a[8] = a[1] + 7d (since the difference between the 8th term and the 1st term is 7 times 'd')
Substituting these expressions into our equation, we get:
a[1] + 3d = 2 * (a[1] + 7d)
Simplifying this equation, we find:
a[1] + 3d = 2a[1] + 14d
a[1] - 2a[1] = 14d - 3d
-a[1] = 11d
Dividing both sides by (-1), we have:
a[1] = -11d
Now, let's find the 10th term, denoted as 'a[10]':
a[10] = a[1] + 9d (since the difference between the 10th term and the 1st term is 9 times 'd')
Substituting the value of 'a[1]' we found, we get:
a[10] = -11d + 9d
a[10] = -2d
Finally, let's find the 11th term, denoted as 'a[11]':
a[11] = a[1] + 10d (since the difference between the 11th term and the 1st term is 10 times 'd')
Substituting the value of 'a[1]' we found, we get:
a[11] = -11d + 10d
a[11] = -d
Now, if we double the equation for the 11th term:
2 * a[11] = -2d
We can see that the double of the 11th term is equal to the 10th term:
2 * a[11] = a[10]
Therefore, we have proved that the 10th term is indeed twice the 11th term in the given arithmetic progression. Keep the jokes aside, math can be fun too!
To prove that the 10th term is twice the 11th term in an arithmetic progression (AP) where the 4th term is twice the 8th term, we'll use the general formula for the nth term of an AP:
aₙ = a₁ + (n-1)d
Where aₙ represents the nth term, a₁ is the first term, n is the position of the term, and d is the common difference.
Given that the 4th term (a₄) is twice the 8th term (a₈), we can write it as an equation:
a₄ = 2a₈
Plugging in the formula for both terms:
a₁ + (4-1)d = 2(a₁ + (8-1)d)
Simplifying the equation:
a₁ + 3d = 2a₁ + 14d
Rearranging terms:
-a₁ = 11d
Dividing both sides by -1:
a₁ = -11d
Now, let's find the 10th term (a₁₀) using the formula:
a₁₀ = a₁ + (10-1)d
Substituting the value we found for a₁:
a₁₀ = -11d + 9d
Simplifying:
a₁₀ = -2d
Similarly, let's find the 11th term (a₁₁):
a₁₁ = a₁ + (11-1)d
Substituting the value we found for a₁:
a₁₁ = -11d + 10d
Simplifying:
a₁₁ = -d
Comparing the values of a₁₀ and a₁₁, we can see that:
a₁₀ = -2d = 2(-d) = 2a₁₁
Therefore, we have proven that the 10th term (a₁₀) in the AP is twice the 11th term (a₁₁).
To prove that the 10th term is twice the 11th term, we first need to understand the given information and the properties of an arithmetic progression (A.P).
An arithmetic progression is a sequence of numbers in which the difference between consecutive terms is constant. In an A.P, the nth term can be represented as:
an = a1 + (n - 1)d
Where:
an = nth term
a1 = first term
n = term number
d = common difference
Given that the 4th term (a4) is twice the 8th term (a8), we can write this as an equation:
a4 = 2 * a8
Using the formula for the nth term, we can obtain the expressions for the 4th and 8th terms:
a4 = a1 + 3d
a8 = a1 + 7d
Substituting these expressions into the equation, we have:
a1 + 3d = 2 * (a1 + 7d)
Now, simplify the equation:
a1 + 3d = 2a1 + 14d
Rearranging the terms, we get:
a1 - 2a1 = 14d - 3d
-a1 = 11d
Dividing both sides by -1, we obtain:
a1 = -11d
Now, let's find the expressions for the 10th (a10) and 11th (a11) terms:
a10 = a1 + 9d
a11 = a1 + 10d
Substituting the value of a1 from above, we have:
a10 = -11d + 9d
a11 = -11d + 10d
Simplifying these expressions, we get:
a10 = -2d
a11 = -d
Therefore, we have successfully shown that the 10th term (a10) is twice the 11th term (a11):
a10 = -2d = 2 * (-d) = 2a11
Hence, we have proved that the 10th term is indeed twice the 11th term in this arithmetic progression.