if the sixth term of an arithmetic progression is 37 and sum of the first six term is 147
the sixth term of an arithmetic progression is 37
---> a+5d = 37
5d = 37-a
sum of the first six term is 147
---> (6/2)(2a + 5d) = 147
3(2a + 37-a) = 147
a +37 = 49
a = 12
d = 5
If you had a question, take it from there
a1 is the initial term of an arithmetic progression.
The common difference of successive members is d.
The nth term of an arithmetic progression.
an = a1 + ( n - 1 ) d
In this case:
n = 6
a6 = a1 + ( 6 - 1 ) d = 37
a1 + 5 d = 37
The sum of the n terms of an arithmetic is:
Sn = n * ( a1 + an ) / 2
S6 = 6 * ( a1 + a6 ) / 2 = 147
3 * ( a1 + a6 ) = 147
3 * ( a1 + a1 + 5 d ) = 147
3 * ( 2 a1 + 5 d ) = 147
6 a1 + 15 d = 147
Now you must solve system:
a1 + 5 d = 37
6 a1 + 15 d = 147
Try that.
The solutions are:
a1 = 12 , d = 5
Proof:
a1 = 12
a2 = 12 + 5 = 17
a3 = 17 + 5 = 22
a4 = 22 + 5 = 27
a5 = 27 + 5 = 32
a6 = 32 + 5 = 37
S6 = a1 + a2 + a3 + a4 + a5 + a6 =
12 + 17 + 22 + 27 + 32 + 37 = 147
What may be the appropriate famular for this question?
To find the difference between consecutive terms in an arithmetic progression, we need to subtract the fifth term from the sixth term.
Using the given information, we know that the sixth term is 37.
To find the fifth term, we need to subtract the common difference from the sixth term. However, since we don't know the common difference yet, let's call it 'd'.
So, the fifth term is 37 - d.
The sum of the first six terms of an arithmetic progression can be calculated using the formula: S = (n/2)(2a + (n-1)d), where S is the sum, n is the number of terms, a is the first term, and d is the common difference.
In this case, we know that the sum of the first six terms is 147. We also know that the first term is a, the common difference is d, and the number of terms is 6.
Substituting the given values into the formula, we get:
147 = (6/2)(2a + (6-1)d)
147 = 3(2a + 5d)
Now, we have two equations:
1) 5th term = 37 - d
2) 147 = 3(2a + 5d)
To solve these equations simultaneously, we can substitute the value of the 5th term from equation 1) into equation 2):
147 = 3(2a + 5(37-d))
147 = 3(2a + 185 - 5d)
147 = 6a + 555 - 15d
6a - 15d = -408 (Subtracting 555 from both sides)
We now have a system of two equations:
1) 5th term = 37 - d
2) 6a - 15d = -408
To solve the system, we can substitute the value of the 5th term from equation 1) into equation 2) and solve for 'a':
6a - 15(37 - d) = -408
6a - 555 + 15d = -408
6a + 15d = 147 (Adding 555 to both sides)
Dividing each term of this equation by 3, we get:
2a + 5d = 49
Now we have a new system of equations:
3) 5th term = 37 - d
4) 2a + 5d = 49
To solve this system, we can substitute the value of the 5th term from equation 3) into equation 4) and solve for 'a':
2a + 5(37 - d) = 49
2a + 185 - 5d = 49
2a - 5d = -136 (Subtracting 185 from both sides)
Now solve equations 2a + 5d = 49 and 2a - 5d = -136 simultaneously:
Adding the two equations, we get:
(2a + 5d) + (2a - 5d) = 49 + (-136)
4a = -87
Dividing by 4, we find:
a = -21.75
Now, substitute the value of 'a' into equation 4):
2(-21.75) + 5d = 49
-43.5 + 5d = 49
5d = 92.5
d = 18.5
The common difference (d) is 18.5, and the first term (a) is -21.75.
Therefore, the arithmetic progression is: -21.75, -3.25, 15.25, 33.75, 52.25, 70.75