The fourth term of an AP is 8 and the sum of the first ten terms is 40. Find the first term and the tenth term.
T4 = a+3d = 8
S10 = 5(2a+9d) = 40
a = 16
d = -8/3
T1 = 16
T10 = 16 + 9(-8/3) = -8
check, sequence is
16 40/3 32/3 8 16/3 8/3 0 -8/3 -16/3 -8 ...
Sum of 1st 10 terms = 40
Why did the arithmetic progression join the circus? Because it wanted to show off its amazing ability to entertain with its perfectly balanced sequence of numbers! Now, let's solve this math puzzle.
Let's denote the first term of the arithmetic progression as 'a' and the common difference as 'd'.
From the information given, we know that the fourth term is 8:
a + 3d = 8 ---(1)
We also know that the sum of the first ten terms is 40:
(10/2)(2a + 9d) = 40 ---(2)
Let's simplify equation (2):
5(2a + 9d) = 40
2a + 9d = 8 ---(3)
Now, we have two equations (1) and (3) with two variables (a and d). Time to solve!
Solving equations (1) and (3) simultaneously, we get:
a = 4
d = 2/3
So, the first term (a) is 4 and the common difference (d) is 2/3.
To find the tenth term, we can substitute the values back into equation (1):
a + 9d = 4 + 9(2/3) = 4 + 6 = 10
So, Mr. First Term is 4 and Mr. Tenth Term is 10! Enjoy the show, and let the arithmetic progression's numbers dance to the rhythm of humor!
To find the first term and the tenth term of an arithmetic progression (AP), we can use the formulas:
nth term = a + (n - 1)d
Sum of n terms = (n/2)(2a + (n - 1)d)
where:
a = first term
n = number of terms
d = common difference
Given:
The fourth term (n = 4) is 8, so we have:
8 = a + (4 - 1)d
8 = a + 3d ---(equation 1)
The sum of the first ten terms (n = 10) is 40, so we have:
40 = (10/2)(2a + (10 - 1)d)
40 = 5(2a + 9d)
40 = 10a + 45d ---(equation 2)
Now, we can solve this system of equations (equation 1 and equation 2) to find the values of 'a' and 'd'.
Multiplying equation 1 by 10, we get:
80 = 10a + 30d ---(equation 3)
Subtracting equation 3 from equation 2, we eliminate 'a' and solve for 'd':
(10a + 45d) - (10a + 30d) = 40 - 80
15d = -40
d = -40/15
d = -8/3
Substituting the value of 'd' = -8/3 back into equation 1, we can solve for 'a':
8 = a + 3(-8/3)
8 = a - 8
a = 8 + 8
a = 16
Therefore, the first term (a) of the arithmetic progression is 16, and the common difference (d) is -8/3.
To find the tenth term, we can use the nth term formula:
t10 = a + (10 - 1)d
t10 = 16 + 9(-8/3)
t10 = 16 - 72/3
t10 = 16 - 24
t10 = -8
Therefore, the tenth term of the arithmetic progression is -8.
To find the first term and the tenth term of an arithmetic progression (AP), we can use the formulas for the nth term and the sum of the first n terms of an AP.
Let's denote the first term as 'a' and the common difference as 'd' (since it's not given in the question).
Given that the fourth term is 8, we can use the formula for the nth term of an AP:
a + (n - 1)d = term
Substituting n = 4 and the given term = 8, we get:
a + (4 - 1)d = 8
a + 3d = 8 ----(equation 1)
Also, given that the sum of the first ten terms is 40, we can use the formula for the sum of the first n terms of an AP:
n/2 * (2a + (n - 1)d) = sum
Substituting n = 10 and the given sum = 40, we get:
10/2 * (2a + (10 - 1)d) = 40
5 * (2a + 9d) = 40
2a + 9d = 8 ----(equation 2)
Now we have two equations (equation 1 and equation 2) in terms of 'a' and 'd'.
We can solve these equations simultaneously to find the values of 'a' and 'd'.
Subtracting equation 1 from equation 2, we get:
2a + 9d - (a + 3d) = 8 - 8
a + 6d = 0
Now, we have the value of 'a' in terms of 'd': a = -6d.
Let's substitute this back into equation 1:
-6d + 3d = 8
-3d = 8
Dividing both sides by -3, we get:
d = -8/3
Now, we can substitute the value of 'd' back into the equation a = -6d:
a = -6 * (-8/3)
a = 16/3
Therefore, the first term (a) is 16/3, and the common difference (d) is -8/3.
To find the tenth term, we use the formula for the nth term again:
a + (n - 1)d = term
Substituting n = 10, a = 16/3, and d = -8/3, we get:
(16/3) + (10 - 1)(-8/3) = term
(16/3) + 9(-8/3) = term
(16/3) - 72/3 = term
-56/3 = term
Therefore, the tenth term is -56/3.
To summarize:
- The first term (a) is 16/3.
- The tenth term is -56/3.