what is the first term of an arithmetic sequence if the 7th term is 21 and the 10th term is 126
7th term is 21 ----> a+6d = 21
10th term is 126 ---> a + 9d = 126
subtract them:
3d = 105
d = 35
back in a+6d = 21 -----> a = -189
first term is -189
check:
t7 = -189 + 6(35) = 21
t10 = -189 + 9(35) = 126
For an arithmetic sequence, the nth term is given by:
T(n)=T(0)+kn,
where k is a constant.
Knowing T(7)=21, and T(10)=126
we find k using
T(10)-T(7) = 126-21
T(0)+10k - (T(0)+7k) = 105
3k = 105
k=35
From the value of k, we find T(0):
T(n)=T(0)+35n
T(7)=T(0)+35(7) = 21
T(0)=21-245=-224
T(n) = -224 + 35n
To find the first term of an arithmetic sequence, we need to use the formula for the nth term of an arithmetic sequence.
The formula for the nth term of an arithmetic sequence is:
an = a1 + (n - 1)d
where:
an is the nth term of the sequence,
a1 is the first term of the sequence,
n is the position of the term in the sequence, and
d is the common difference between the terms.
Given that the 7th term is 21 and the 10th term is 126:
a7 = a1 + (7 - 1)d
21 = a1 + 6d
a10 = a1 + (10 - 1)d
126 = a1 + 9d
We have a system of two equations with two variables. We can solve this system of equations to find the values of a1 and d.
Subtract the first equation from the second equation:
126 - 21 = a1 + 9d - a1 - 6d
105 = 3d
Divide both sides by 3:
105 / 3 = 3d / 3
35 = d
Now substitute the value of d back into the first equation:
21 = a1 + 6(35)
21 = a1 + 210
Subtract 210 from both sides:
21 - 210 = a1 + 210 - 210
-189 = a1
Therefore, the first term of the arithmetic sequence is -189.
To find the first term of an arithmetic sequence, we need to determine the common difference.
We can start by calculating the common difference using the formula for the nth term of an arithmetic sequence:
an = a1 + (n - 1)d
Where:
- an represents the nth term
- a1 represents the first term
- d represents the common difference
Given that the 7th term (a7) is 21 and the 10th term (a10) is 126, we can plug these values into the formula:
a7 = a1 + (7 - 1)d
21 = a1 + 6d -> Equation 1
a10 = a1 + (10 - 1)d
126 = a1 + 9d -> Equation 2
Next, we need to solve these two equations simultaneously to find the values of a1 and d.
Multiplying Equation 1 by 9 and Equation 2 by 6, we get:
189 = 9a1 + 54d -> Equation 3
126 = 6a1 + 54d -> Equation 4
Subtracting Equation 4 from Equation 3, we eliminate the d variable:
189 - 126 = 9a1 - 6a1
63 = 3a1
Dividing both sides by 3, we find:
a1 = 63 / 3
a1 = 21
Therefore, the first term (a1) of the arithmetic sequence is 21.