Two APs have the same first and last terms. The first AP has 21 terms with a common difference of 9. How many terms has the other AP if it's common difference is 4?
The formula for the nth term of an arithmetic progression is given by:
An = A1 + (n-1)d,
where An is the nth term, A1 is the first term, n is the number of terms, and d is the common difference.
For the first AP:
A1 = first term = A21 = last term
n = 21
d1 = 9
Using the formula, we can find A1:
A1 = A21 - (21-1)d1
A1 = A21 - 20d1
For the second AP:
d2 = 4
Using the formula, we can find the number of terms (n2) in the second AP:
A1 = A2 - (2-1)d2
A1 = A2 - d2
Since A1 is equal to the last term of the first AP (A1 = A21), we can substitute A21 - 20d1 for A1:
A21 - 20d1 = A2 - d2
Simplifying, we get:
21 - 20(9) = A2 - 4
21 - 180 = A2 - 4
A2 - 4 = -159
Thus, the number of terms in the second AP is 159.
To solve this problem, we can use the formula for the nth term of an arithmetic sequence given by:
nth term = first term + (n - 1) * common difference
In the first arithmetic sequence (AP), we are given that it has 21 terms with a common difference of 9. Let's calculate the first and last terms of this AP:
First term = a
Common difference = d
Number of terms = n
First term (a) = ?
Common difference (d) = 9
Number of terms (n) = 21
Using the formula, we can write the equation for the nth term as:
a + (n - 1) * d = last term
Substituting the known values, we have:
a + (21 - 1) * 9 = last term
Simplifying further:
a + 20 * 9 = last term
a + 180 = last term
Since the first and last terms are the same in both APs, we can equate the last term of the first AP to the first term of the second AP:
last term of first AP = first term of second AP
a + 180 = first term of second AP
Now, let's find the number of terms in the second arithmetic sequence (AP) with a common difference of 4.
Common difference (d) of second AP = 4
Number of terms (n) in the second AP = ?
Using the formula for the nth term, we can write the equation:
(a + 180) + (n - 1) * 4 = first term of second AP + (n - 1) * 4
To solve for n, we need to find the values of a (first term of second AP) and the last term of the second AP.
From the equation above, we can see that the last term of the second AP is:
Last term of second AP = (a + 180) + (n - 1) * 4
Since the first and last terms are the same in both APs, we can equate the last term of the first AP to the last term of the second AP:
last term of first AP = last term of second AP
a + 180 = (a + 180) + (n - 1) * 4
By canceling out the common terms, we can solve for n:
0 = (n - 1) * 4
Now, let's solve for n:
0 = 4n - 4
4 = 4n
n = 1
Therefore, the second AP has 1 term with a common difference of 4.
To find the number of terms in the second arithmetic progression (AP) with a common difference of 4, we need to determine the number of terms in the first AP and its common difference.
In the first AP, we are given that there are 21 terms and the common difference is 9. Using these values, we can calculate the last term of the first AP.
The formula to find the nth term of an AP is given by:
a_n = a_1 + (n - 1) * d,
where a_n represents the nth term, a_1 is the first term, n is the number of terms, and d is the common difference.
We can substitute the given values for the first AP into the formula:
a_21 = a_1 + (21 - 1) * 9.
Simplifying this equation:
a_21 = a_1 + 20 * 9,
a_21 = a_1 + 180.
Since we are told that the first AP has the same first and last terms, we can also conclude that the first term of the first AP is equal to the last term. Therefore, we can substitute a_1 with a_n to simplify the equation further:
a_n = a_n + 180.
Subtracting a_n from both sides of the equation, we obtain:
0 = 180.
This equation does not have a valid solution. Therefore, there is no valid second AP that satisfies the given condition, as it would imply that the sum of the first AP is equal to zero, which is not possible in this case.