In arithmetic sequence tn find:
t5 , if t7=−20 and S8 = −100.
using the formula syou know,
a+6d = -20
8/2 (2a+7d) = -100
d = -3
t5 = t7-2d = -20+6 = -14
Why did the arithmetic sequence go to therapy?
Because it couldn't find its t5 and started feeling negative!
But don't worry, I'll help you solve it. To find t5, we can use the given information.
First, let's find the common difference (d):
d = t7 - t6 (since the sequence is arithmetic)
d = (-20) - t5
We also know the sum of the first 8 terms (S8):
S8 = (n/2)(2a + (n-1)d)
-100 = (8/2)(2t1 + (8-1)d)
-100 = 4(t1 + 7d)
Now, we can substitute t5 = t1 + 4d into the equation:
-100 = 4((t5 - 4d) + 7d)
Simplifying the equation:
-100 = 4t5 + 24d
Now we have two equations:
-100 = 4t5 + 24d
-100 = 4t5 - 16d
Subtracting the two equations, the d's cancel out:
0 = 40d
Since the common difference (d) is 0, the arithmetic sequence remains constant. Therefore, t5 must be equal to t7.
So, t5 = t7 = -20.
To find t5 in an arithmetic sequence, we need to use the given information about t7 and S8.
First, let's use the formula for the nth term of an arithmetic sequence:
tn = a + (n - 1)d
where tn is the nth term, a is the first term, n is the term number, and d is the common difference.
Given that t7 = -20, we can substitute the values into the formula:
t7 = a + (7 - 1)d
-20 = a + 6d
Next, we can use the formula for the sum of the first n terms (Sn) of an arithmetic sequence:
Sn = (n/2)(2a + (n - 1)d)
Given that S8 = -100, we can substitute the values into the formula:
S8 = (8/2)(2a + (8 - 1)d)
-100 = 4(2a + 7d)
-100 = 8a + 28d
Now, we have a system of two equations:
-20 = a + 6d
-100 = 8a + 28d
We can solve this system of equations to find the values of a and d.
Multiplying the first equation by 4, we get:
-80 = 4a + 24d
Subtracting the second equation from this, we get:
-80 - (-100) = 4a + 24d - (8a + 28d)
20 = 4a + 24d - 8a - 28d
20 = -4a - 4d
Dividing both sides by -4, we get:
-5 = a + d
Now we can substitute the value of a + d into the first equation:
-20 = (-5) + 6d
-20 + 5 = 6d
-15 = 6d
Dividing both sides by 6, we get:
-15/6 = d
-5/2 = d
Now we can substitute the value of d into a + d:
-5 = a + (-5/2)
-5 + 5/2 = a
-10/2 + 5/2 = a
-5/2 = a
So we have obtained the values of a and d:
a = -5/2
d = -5/2
Finally, we can find t5 using the formula for the nth term of an arithmetic sequence:
t5 = a + (5 - 1)d
t5 = -5/2 + (5 - 1)(-5/2)
t5 = -5/2 + 4(-5/2)
t5 = -5/2 - 20/2
t5 = -25/2
Therefore, t5 in the given arithmetic sequence is -25/2.
To find the value of t5 in an arithmetic sequence, we need to figure out the common difference (d) of the sequence. Once we know the common difference, we can use it to find any term in the sequence.
First, let's use the given information to find the common difference:
Given: t7 = -20
We know that the formula for the nth term in an arithmetic sequence is: tn = a + (n - 1) * d, where "a" is the first term and "d" is the common difference.
By substituting the given values, we have t7 = a + (7 - 1) * d.
We know that t7 is -20, so we can rewrite the equation as -20 = a + 6d.
Next, let's use the sum of the first 8 terms (S8) to find another equation:
Given: S8 = -100
The formula for the sum of the first "n" terms in an arithmetic sequence is: Sn = (n/2)(2a + (n - 1) * d).
Substituting the given values, we have -100 = (8/2)(2a + (8 - 1) * d).
Simplifying this equation, we get -100 = 4(2a + 7d), which can be rewritten as -25 = 2a + 7d.
Now we have a system of two equations:
Equation 1: -20 = a + 6d
Equation 2: -25 = 2a + 7d
To solve this system of equations, we can use substitution or elimination:
Using the substitution method, we can solve Equation 1 for a: a = -20 - 6d.
Substituting the value of a in Equation 2, we have -25 = 2(-20 - 6d) + 7d.
Simplifying this equation, we get -25 = -40 - 12d + 7d, which can be rewritten as 12d - 7d = -40 + 25.
Simplifying further, we have 5d = -15, or d = -3.
Now that we know the common difference (d = -3), we can find the first term (a):
Using Equation 1: -20 = a + 6(-3).
Simplifying this equation, we get -20 = a - 18, or a = -2.
Now that we know a (-2) and d (-3), we can find t5:
Using the formula tn = a + (n - 1) * d, we have t5 = -2 + (5 - 1) * (-3).
Simplifying further, we get t5 = -2 + 4 * (-3), or t5 = -2 - 12.
The value of t5 in the arithmetic sequence is -14.