The 12th term of exceeds the 3rd term by 36 of an a.p
If it's 16th term is 64 find first term and common difference
a+11d - (a + 2d) = 36
9d = 36
d = 4
term(16) = a+15d = 64
a + 60 = 64
a = 4
so a = 4, d = 4
Very easy method
Well, well, well! Looks like we've got ourselves an arithmetic progression problem. Don't worry, I'm here to clown around and help you out!
Let's start by finding the difference between the 12th and 3rd terms. You say it's 36? No problemo! We can work with that.
The formula to find the nth term of an arithmetic progression is Tn = a + (n - 1)d, where Tn is the nth term, a is the first term, and d is the common difference.
Since the 12th term exceeds the 3rd term by 36, we can set up an equation: T12 - T3 = 36. Substituting the formula, we get:
(a + 11d) - (a + 2d) = 36.
Let's simplify that: 9d = 36. Divide both sides by 9, and we find that d = 4.
Now we know the common difference is 4. Easy-peasy, right?
To find the first term, we can use the 16th term. You mentioned it's 64, right? Let's use our formula again:
T16 = a + (16 - 1)4. Simplifying further, we get:
T16 = a + 60.
Since we know the 16th term is 64, we can substitute that in: 64 = a + 60. Subtraction time!
4 = a.
So, the first term is 4 and the common difference is 4 too. Ta-da! Problem solved, and we've had some fun along the way. Keep clownin' around with math, my friend!
Let's denote the first term as 'a' and the common difference as 'd'.
We are given that the 12th term exceeds the 3rd term by 36. This can be expressed as:
a + 11d = a + 2d + 36 ----(1)
We are also given that the 16th term, which is (a + 15d), is equal to 64:
a + 15d = 64 ----(2)
We can solve these two equations to find the values of 'a' and 'd'.
Subtracting equation (1) from equation (2), we get:
(a + 15d) - (a + 11d) = 64 - (a + 2d + 36)
4d = -a + 2d + 28
2d = -a + 28 ----(3)
Now, we substitute equation (3) into equation (1) to eliminate 'a':
a + 11d = -a + 2d + 36
2a + 9d = 36 ----(4)
To solve equations (3) and (4), we can use either substitution or elimination method. Let's use the substitution method.
From equation (3), we can express 'a' in terms of 'd':
a = 28 - 2d
Substituting this value of 'a' into equation (4), we get:
2(28 - 2d) + 9d = 36
56 - 4d + 9d = 36
5d = 36 - 56
5d = -20
d = -20/5
d = -4
Now that we have found the common difference 'd', we can substitute this value into equation (3) to find 'a':
a = 28 - 2d
a = 28 - 2(-4)
a = 28 + 8
a = 36
Therefore, the first term (a) is 36 and the common difference (d) is -4.
To solve this question, we need to use the formula for the nth term of an arithmetic progression (AP):
\[a_n = a_1 + (n-1)d\]
where \(a_n\) represents the \(n\)th term, \(a_1\) is the first term, \(n\) is the number of terms, and \(d\) is the common difference.
Given information:
The 12th term exceeds the 3rd term by 36, so we can write this as:
\[a_{12} = a_3 + 36\]
The 16th term is 64:
\[a_{16} = 64\]
We need to find the first term (\(a_1\)) and the common difference (\(d\)).
To solve for \(a_1\) and \(d\), we can use a system of equations. First, we substitute the values in the formula:
For the 12th term:
\[a_12 = a_1 + (12-1)d\]
For the 3rd term:
\[a_3 = a_1 + (3-1)d\]
Now we can substitute these into the equation we obtained earlier:
\[a_{12} = a_3 + 36\]
\[(a_1 + (12-1)d) = (a_1 + (3-1)d) + 36\]
Simplifying, we get:
\[a_1 + 11d = a_1 + 2d + 36\]
\[9d = 36\]
\[d = 4\]
Substituting this value of \(d\) back into the equation for the 3rd term:
\[a_3 = a_1 + (3-1)(4)\]
\[a_3 = a_1 + 2(4)\]
\[a_3 = a_1 + 8\]
Now we substitute the value for the 3rd term and solves for \(a_1\) using the equation for the 12th term:
\[a_{12} = a_3 + 36\]
\[a_1 + (12-1)(4) = (a_1 + 8) + 36\]
\[a_1 + 11(4) = a_1 + 8 + 36\]
\[a_1 + 44 = a_1 + 44\]
\[a_1 = 0\]
Therefore, the first term (\(a_1\)) of the arithmetic progression is 0 and the common difference (\(d\)) is 4.