Find the 6th term of the arithmetic sequence with a9=120 and a14=195
a9 = 120 ----> a + 8d = 120
a14 = 195 ---> a + 13d = 195
subtract them
5d = 75
d = 15
then a+8(15) = 120
a = 0
so term(6) = a+5d = 0 + 5(15) = 75
To find the 6th term of the arithmetic sequence, we need to first find the common difference (d).
Using the formula for arithmetic sequence:
a_n = a_1 + (n-1)d
We have two equations:
a_9 = a_1 + 8d = 120 ...........(1)
a_14 = a_1 + 13d = 195 ...........(2)
Subtracting equation (1) from equation (2), we get:
a_14 - a_9 = (a_1 + 13d) - (a_1 + 8d)
195 - 120 = 5d
75 = 5d
d = 15
Now that we have the common difference (d = 15), we can find the 6th term (a_6) using equation (1):
a_6 = a_1 + (6-1)d
a_6 = a_1 + 5d
a_6 = a_1 + 5(15)
a_6 = a_1 + 75
However, we don't have the value of a_1 given in the question. Hence, we cannot find the 6th term of the arithmetic sequence without knowing the value of a_1.
To find the 6th term of an arithmetic sequence, we need to know the first term (a1) and the common difference (d).
To find the common difference, we need to find the difference between any two consecutive terms. In this case, we have a9=120 and a14=195.
The formula to find the nth term of an arithmetic sequence is:
an = a1 + (n-1)d
We can use this formula to find the common difference:
a9 = a1 + (9-1)d
120 = a1 + 8d
a14 = a1 + (14-1)d
195 = a1 + 13d
Now we have a system of two equations:
a1 + 8d = 120
a1 + 13d = 195
We can solve this system of equations to find the values of a1 and d.
Subtract the first equation from the second equation:
(13d - 8d) = (195 - 120)
5d = 75
d = 15
Now, substitute the value of d back into the first equation to find a1:
a1 + 8(15) = 120
a1 + 120 = 120
a1 = 120 - 120
a1 = 0
So, we have found that a1 = 0 and d = 15.
Now, substitute the values of a1 and d into the formula to find the 6th term (a6):
a6 = a1 + (6-1)d
a6 = 0 + (5)(15)
a6 = 0 + 75
a6 = 75
Therefore, the 6th term of the arithmetic sequence is 75.