in an arthmetic progression or series,the 13th term is 27 and the 7th iterm is 3 times the 2nd term ,find first term the common difference the sum of rhe squence
13th term is 27 --- a + 12d = 27
the 7th iterm is 3 times the 2nd term --> a+6d = 3(a+d)
or
a+6d = 3a + 3d
2a = 3d
a = 3d/2
sub into the first equation
3d/2 + 12d = 27
3d+24d=54
d = 2
then a = 3(2)/2 = 3
I will let you finish it for the sum, you did not say how many terms you want to add.
To find the first term, common difference, and sum of the arithmetic progression, we can use the given information.
Let's denote the first term as 'a' and the common difference as 'd'.
We are given two pieces of information:
1. The 13th term is 27, which means a + 12d = 27.
2. The 7th term is 3 times the 2nd term, which means a + 6d = 3(a + d).
We have two equations:
Equation 1: a + 6d = 3(a + d)
Equation 2: a + 12d = 27
To solve these equations, we can use the method of substitution or elimination.
Let's solve the equations using substitution:
From Equation 1, we can rewrite it as:
a = 3(a + d) - 6d
a = 3a + 3d - 6d
a = 3a - 3d
Adding -3a to both sides:
0 = -3d
Dividing both sides by -3:
0/d = -3d/-3
0 = d
Since the common difference 'd' is zero, this means it is not an arithmetic progression but rather a sequence of the same repeating number.
Now, let's find the first term and the sum of the sequence:
Using Equation 2, we substitute the value of 'd' (0) into it:
a + 12(0) = 27
a = 27
Therefore, the first term is 27, the common difference is 0, and the sum of the sequence is simply the value of the first term since all terms are the same.
So, the first term is 27, the common difference is 0, and the sum of the sequence is also 27.