The 5th term of an arithmetic progression is 23 and the 12th term is 37.find the third term and common different
Just use your definitions:
"5th term of an arithmetic progression is 23"
---> a + 4d =23
"the 12th term is 37" --> a + 11d = 37
You have two equations in two unknows.
Solve them.
Hint: I would subtract one from the other.
Let me know what you get.
a + 4d=23
a + 11d=37
a + 7d=14
a +7dรท7 =14รท7
1a + d =2
To find the third term and common difference of an arithmetic progression, we can use the formula:
๐๐ = ๐โ + (๐ โ 1)๐
where:
๐โ is the ๐-th term of the sequence
๐โ is the first term of the sequence
๐ is the position of the term in the sequence
๐ is the common difference
Given that the 5th term is 23 and the 12th term is 37, we can use this information to form two equations and solve them simultaneously to find ๐โ and ๐.
Equation 1: ๐โ
= ๐โ + 4๐ = 23
Equation 2: ๐โโ = ๐โ + 11๐ = 37
To solve these equations, we can use the method of substitution or elimination. Let's use the method of substitution:
From Equation 1, we can express ๐โ in terms of ๐:
๐โ = 23 - 4๐
Substituting this back into Equation 2:
23 - 4๐ + 11๐ = 37
Simplifying the equation:
23 + 7๐ = 37
Subtracting 23 from both sides:
7๐ = 14
Dividing both sides by 7:
๐ = 2
We have found the common difference, which is 2.
Now, substituting this value of ๐ into Equation 1 to find ๐โ:
๐โ = 23 - 4(2) = 23 - 8 = 15
Therefore, the first term ๐โ is 15, and the common difference ๐ is 2.
To find the third term, we use ๐โ = ๐โ + 2๐, where ๐ is the common difference. Substituting the values we have found:
๐โ = 15 + 2(2) = 15 + 4 = 19
So, the third term of the arithmetic progression is 19.
To find the third term and the common difference of an arithmetic progression, we can use the following formula:
An = A1 + (n-1)d
Where:
An is the nth term
A1 is the first term
d is the common difference
n is the term number
Given that the 5th term is 23, we can substitute the values into the formula:
23 = A1 + (5-1)d
Simplifying the equation:
23 = A1 + 4d
Similarly, given that the 12th term is 37, we can substitute the values:
37 = A1 + (12-1)d
37 = A1 + 11d
We now have two equations with two unknowns (A1 and d). We can solve these equations simultaneously to find the values of the unknowns.
Subtracting the first equation from the second equation, we get:
37 - 23 = A1 + 11d - (A1 + 4d)
14 = 11d - 4d
14 = 7d
Dividing both sides by 7:
d = 2
Now that we have found the common difference (d = 2), we can substitute it back into either equation to find the value of A1.
Using the first equation:
23 = A1 + 4(2)
23 = A1 + 8
Subtracting 8 from both sides:
15 = A1
Hence, the third term (A3) is given by:
A3 = A1 + (3-1)d
A3 = 15 + (3-1)2
A3 = 15 + 2(2)
A3 = 15 + 4
A3 = 19
Therefore, the third term is 19 and the common difference is 2.