The 5th and 10th terms of a linear sequence are -12 and -27 respectively.Find first term and the common difference of the sequence
use your definitions:
a+4d = -12
a+9d = -27
subtract them,
5d = -15
d = -3
back into a+4d = -12
a - 12 = -12
a = 0
check: sequence is 0,-3,-6,-9,-12,-15,-18,-21,-24,-27,-30
yup, we got it
Linear sequence is other name for arithmetic sequence.
For arithmetic sequence nth term:
an = a1 + ( n - 1 ) d
where:
a1 = initial term of arithmetic progression
d = common difference
In this case:
a5 = a1 + ( 5 - 1 ) d
a5 = a1 + 4 d
a10 = a1 + ( 10 - 1 ) d
a10 = a1 + 9 d
so:
a5 = -12
- 12 = a1 + 4 d
a10 = - 27
- 27 = a1 + 9 d
Now you must solve system:
- 12 = a1 + 4 d
- 27 = a1 + 9 d
Try it.
The solutions are:
a1 = 0
d = - 3
Your sequence:
0 , - 3 , - 6 , - 9 , - 12 , -15 , - 18 , - 21 , - 24 , - 27...
To find the first term and the common difference of a linear sequence, we can use the formula:
an = a1 + (n-1)d
where:
an is the nth term
a1 is the first term
d is the common difference
n is the position of the term in the sequence
Given that the 5th term (n=5) is -12 and the 10th term (n=10) is -27, we have:
a5 = a1 + (5-1)d = -12
a10 = a1 + (10-1)d = -27
Now we can create a system of equations:
a1 + 4d = -12 // from a5 = -12
a1 + 9d = -27 // from a10 = -27
To solve the system of equations, we can subtract the first equation from the second equation to eliminate a1:
(a1 + 9d) - (a1 + 4d) = -27 - (-12)
9d - 4d = -27 + 12
5d = -15
d = -15/5
d = -3
Now we can substitute d = -3 back into either equation to find a1:
a1 + 4(-3) = -12
a1 - 12 = -12
a1 = 0
Therefore, the first term (a1) of the sequence is 0 and the common difference (d) is -3.
To find the first term (a₁) and the common difference (d) of a linear sequence, we can use the formula for the nth term of a linear sequence, which is:
an = a₁ + (n - 1) * d
Given that the 5th term (a₅) is -12 and the 10th term (a₁₀) is -27, we can substitute these values into the formula to create two equations:
-12 = a₁ + (5 - 1) * d
-27 = a₁ + (10 - 1) * d
Now, let's solve these equations to find the values of a₁ and d.
For the first equation, we can simplify it:
-12 = a₁ + 4d (equation 1)
For the second equation, we can simplify it as well:
-27 = a₁ + 9d (equation 2)
Now, we have a system of two linear equations. We can solve this system by subtracting equation 1 from equation 2 to eliminate a₁:
-27 - (-12) = (a₁ + 9d) - (a₁ + 4d)
-27 + 12 = a₁ + 9d - a₁ - 4d
-15 = 5d
Next, we can solve for d:
-15 / 5 = d
-3 = d
Now that we know the common difference (d = -3), we can substitute this back into either of the original equations to find the first term:
-12 = a₁ + 4d
-12 = a₁ + 4(-3)
-12 = a₁ - 12
a₁ = 0
Therefore, the first term (a₁) of the sequence is 0, and the common difference (d) is -3.