the first third and ninth term of a linear sequence are the first three terms of an exponential sequence.if the 7th term of the linear sequence is 14 calcuate the 20th term of the linear sequnce
How do "u solve all that"
the first, third and ni th term of an a.p are the first three term of an expotential sequence (g.p) if the seventh term of the linear equation is 14 calculate: the twentieth term of the sequence, the sum of the first twelve terms of the expotential sequence you need net to simplify any expotential term on your answer, find the sum of the first 101th term of the line sequence (A.P)
Why did the linear sequence get into a relationship with the exponential sequence? Because they heard they made an "exponentially" good pair!
But enough clowning around, let's solve the problem.
If the 7th term of the linear sequence is 14, we can determine the common difference (d) of the linear sequence. Since the linear sequence is given as a7 = 14, we can use the formula for the nth term of a linear sequence:
an = a1 + (n - 1)d
Plugging in the values, we get:
14 = a1 + (7 - 1)d
14 = a1 + 6d
Subtracting 6d from both sides:
14 - 6d = a1
Now, let's find the ratio (r) of the exponential sequence. Since the first, third, and ninth terms of the linear sequence are the first three terms of the exponential sequence, we can use the formula for the nth term of an exponential sequence:
an = a1 * r^(n-1)
From this information, we can deduce the values of a1 and r for the exponential sequence. Let's calculate them:
Using a1 = 14 - 6d, we can say:
a1 = 14 - 6d
Given that the first term of the exponential sequence is a1 (which is also a1 of the linear sequence), we can equate it to the first term of the linear sequence:
a1 = 14 - 6d
Substituting this value into the formula for the nth term of an exponential sequence:
a3 = a1 * r^(3-1)
1st term of exponential seq = (14 - 6d) * r^2
Similarly, for the ninth term:
a9 = a1 * r^(9-1)
9th term of exponential seq = (14 - 6d) * r^8
Now, equating the two expressions, we can find the value of r:
(14 - 6d) * r^2 = (14 - 6d) * r^8
Since (14 - 6d) is nonzero, we can divide both sides by (14 - 6d):
r^2 = r^8
Canceling out the "r" terms:
1 = r^6
Taking the sixth root of both sides:
1^(1/6) = r
Simplifying:
r = 1
Now that we have the values of a1 and r, we can calculate the 20th term of the linear sequence. Using the formula for the nth term of an exponential sequence, where a1 = 14 - 6d and r = 1:
a20 = a1 + (20 - 1)d
a20 = (14 - 6d) + 19d
a20 = 14 + 13d
Therefore, the 20th term of the linear sequence is 14 + 13d, where d is the common difference of the linear sequence.
To find the 20th term of the linear sequence, we need to first determine the common difference between consecutive terms.
Let's represent the linear sequence as an equation: a_n = a + (n-1)d, where a is the first term, d is the common difference, and n is the term number.
Given that the 7th term of the linear sequence is 14, we can substitute the values in the equation: a + 6d = 14. (Equation 1)
Now, let's consider the relationship between the linear and exponential sequences. Since the first, third, and ninth terms of the linear sequence are the first three terms of an exponential sequence, we can express them as: a_1 = a, a_3 = a + 2d, and a_9 = a + 8d.
In an exponential sequence, each term is calculated by multiplying the previous term by a constant ratio (r). Thus, we can write:
a_1 * r^2 = a_3 (Equation 2)
a_1 * r^8 = a_9 (Equation 3)
Substituting the linear sequence expressions into Equations 2 and 3, we get:
a * r^2 = a + 2d (Equation 4)
a * r^8 = a + 8d (Equation 5)
Now, we have four equations (Equations 1, 4, 5, and the equation for the linear sequence) with four variables (a, d, r, and n) that we can solve simultaneously.
From Equation 1: a + 6d = 14
From Equation 4: a * r^2 = a + 2d
From Equation 5: a * r^8 = a + 8d
First, let's solve for a in Equation 1:
a = 14 - 6d
Substituting this value of a into Equations 4 and 5:
(14 - 6d) * r^2 = (14 - 6d) + 2d (Equation 6)
(14 - 6d) * r^8 = (14 - 6d) + 8d (Equation 7)
Now, we have two equations (Equations 6 and 7) with two variables (d and r) that we can solve.
Let's simplify Equation 6:
14r^2 - 6dr^2 = 14 - 4d + 2d
14r^2 - 6dr^2 = 14 - 2d (Equation 8)
Similarly, let's simplify Equation 7:
14r^8 - 6dr^8 = 14 + 2d (Equation 9)
Now, we have two equations (Equations 8 and 9) with two variables (d and r) that we can solve.
From Equation 8: 14r^2 - 6dr^2 = 14 - 2d
From Equation 9: 14r^8 - 6dr^8 = 14 + 2d
Now we can solve these two equations using algebraic methods or numerical methods (such as graphing or numerical approximation).
Once we find the values of d and r, we can substitute them into the equation for the linear sequence: a_n = a + (n-1)d, where n is 20, and find the 20th term of the linear sequence.
AP:
T1 = a
T3 = a+2d
T9 = a+8d
GP:
T1 = a
T2 = ar
T3 = ar^2
a+6d = 14
a+2d = ar
a+8d = ar^2
Solve all that and you find
a=2
d=2
r=3
T20 of the AP is 2+19*2 = 40
Check:
AP: 2 4 6 8 10 12 14 16 18 20 ...
GP: 2 6 18 ...