the 4th term is 37 and the 6th term is 12 more than the 4th term find the first and seventh term
i dint understand
going with arithmetic (not geometric)
two terms apart is two differences ... 2 d = 12 ... d = 6
1st term is 3 d below 37
7th term is 3 d above 37
To find the first and seventh term of a sequence given limited information, you first need to find the common difference.
Let's assume the first term is represented by 'a' and the common difference is represented by 'd'.
Given that the 4th term is 37, we can write the equation:
a + 3d = 37
Given that the 6th term is 12 more than the 4th term, we can write the equation:
a + 5d = 37 + 12
Now we have a system of equations:
a + 3d = 37
a + 5d = 49
To solve this system of equations, we can either use substitution or elimination method.
Using elimination method, we subtract the equations to eliminate 'a':
(a + 5d) - (a + 3d) = 49 - 37
2d = 12
d = 6
Substituting the value of 'd' into any of the original equations, we can solve for 'a':
a + 3(6) = 37
a + 18 = 37
a = 37 - 18
a = 19
Therefore, the first term (7th term) is 19 (a = 19) and the common difference is 6 (d = 6).
To find the 7th term, we substitute the value of 'n=7' into the formula for the nth term of an arithmetic sequence:
a_n = a + (n-1)d
Using 'a = 19', 'n = 7', and 'd = 6', we can calculate the 7th term:
a_7 = 19 + (7-1)6
a_7 = 19 + 6(6)
a_7 = 19 + 36
a_7 = 55
Therefore, the seventh term is 55.