The sum of the first six term of an arithmetic progression is 21 and the 7th term is three time the sum of the third and fourth term find the first term and the common difference.
The sum of the first six term of an arithmetic progression is 21
---- > (6/2)(2a + 5d) = 21
2a + 5d = 7
the 7th term is three time the sum of the third and fourth term
---> a+6d = 3(a+2d + a+3d)
a+6d = 6a + 15d
-5a = 9d
a = -9d/5
sub into 2a + 5d = 7
2(-9d/5) + 5d = 7
times 5
-18d + 25d = 35
7d = 35
d = 5, then a = -9(5)/5 = -9
check:
sum(6) = 3(-18+25) = 21
sum of 3rd and 4th = a + 2d + a + 3d = 2a + 5d = -18 + 25 = 7
7th term = a + 6d = -9 + 30 = 21 , which is 3 times 7
all is ok
Thanks math helper
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Well, let me calculate it while keeping things light and funny!
Let's call the first term of the arithmetic progression "A" and the common difference "D."
The sum of the first six terms can be found using the formula: S = (n/2)(2A + (n-1)D), where "S" is the sum of the first six terms and "n" is the number of terms.
In this case, we have S = 21 and n = 6, so 21 = (6/2)(2A + (6-1)D).
Simplifying further, we get 21 = 3(2A + 5D).
Now, let's focus on the 7th term. It is said to be three times the sum of the third and fourth terms. So, using the same formula, the 7th term can be written as 7th term = (2A + (7-1)D) = 3((2A + (3-1)D) + (2A + (4-1)D)).
Simplifying this, we get (6A + 6D) = 3(4A + 6D).
If we rearrange the equation, it becomes 2A - 6D = 0.
Now we have a system of equations:
Equation 1: 21 = 3(2A + 5D)
Equation 2: 2A - 6D = 0
I'm no mathematician, but I can surely make you smile! So, let me solve this pair of equations for you!
By substituting 2A - 6D = 0 into the first equation, we get:
21 = 3(2(6D) + 5D)
21 = 3(12D + 5D)
21 = 3(17D)
21 = 51D
Well, that's odd! It seems D is not playing by the rules of arithmetic progression! Since 21 divided by 51 isn't a whole number, we can't find a common difference that satisfies the given conditions.
Sorry to be the clown in this situation, but it looks like there might be an error in the problem. Double-check to make sure all the details are accurate!
To find the first term and the common difference of an arithmetic progression, we need to use the given information and apply the formulas for the sum of an arithmetic progression and the general term of an arithmetic progression.
Let's start step by step:
1. The sum of the first six terms of an arithmetic progression is given as 21, which means:
S₆ = 21
We can use the formula for the sum of an arithmetic progression:
Sₙ = (n/2)(2a + (n-1)d)
Here, Sₙ represents the sum of the first n terms, a represents the first term, and d represents the common difference.
Plugging in the given value:
21 = (6/2)(2a + (6-1)d) -> 21 = 3(2a + 5d)
2. The 7th term is three times the sum of the third and fourth terms, which can be represented as:
a + 6d = 3[(a + 2d) + (a + 3d)]
Simplifying:
a + 6d = 3(2a + 5d) -> a + 6d = 6a + 15d
3. Now we have two equations:
21 = 3(2a + 5d) -> equation (1)
a + 6d = 6a + 15d -> equation (2)
We can solve this system of equations to find the values of a (first term) and d (common difference).
To solve this system of equations, we can use any method such as substitution, elimination, or matrices.
Let's proceed by using the substitution method:
From equation (2), we can isolate 'a' in terms of 'd':
a = 6a + 15d - 6d
a = 5a + 9d
4a = 9d
a = (9/4)d
Now substitute this value of 'a' in equation (1):
21 = 3(2(9/4)d + 5d)
21 = 3(9/2)d + 15d
21 = (27/2)d + 15d
21 = (27/2 + 30/2)d
21 = (57/2)d
d = (2/57)*21
d = 14/19
Now substitute the value of 'd' back into the expression for 'a':
a = (9/4)(14/19)
a = 7/2
Therefore, the first term (a) is 7/2 and the common difference (d) is 14/19.