The first, third and night terms of a linear sequence are the first three terms of an exponential sequence G. P if the seventh term of the linear sequence is 14.find the twentieth term of the linear sequence.
for the AS"
first term --- a
third term -- a+2d
ninth term -- a+8d
but those terms form a GP
(a+2d)/a = (a+8d)/(a+2d)
(a+2d)^2 = a(a+8d)
a^2 + 4ad + 4d^2 = a^2 + 8ad
4d^2 - 4ad = 0
d^2 - ad = 9
d(d-a) = 0
d = 0 ---> not too interesting
or
a = d
but term 7 of the AS = 14
a+6d = 14
a + 6a = 14
a = 2, then d = 2
term(20) = a + 19d = 2 + 38 = 40
To find the twentieth term of the linear sequence, we need to determine the common difference of the sequence.
Let's denote the first term of the linear sequence as 'a' and the common difference as 'd'.
Given that the seventh term of the linear sequence is 14, we can write the equation:
a + 6d = 14 (since the seventh term is 'a + 6d').
Since the first, third, and ninth terms of the linear sequence are the first three terms of an exponential sequence G.P, we can write the following equations:
a + 2d = ar (equation 1)
a + 8d = ar^2 (equation 2)
We can solve these equations to find the values of 'a', 'd', and 'r'.
Subtracting equation 1 from equation 2, we get:
6d = ar^2 - ar (equation 3)
Dividing equation 3 by equation 1, we get:
6 = r - 1
Simplifying, we have:
r = 7
Now, we can substitute the value of 'r' into equation 1:
a + 2d = ar
a + 2d = 7a
6a = 2d
Dividing both sides by 2, we have:
3a = d
Now, we can substitute the values of 'r' and 'd' into the equation a + 6d = 14:
a + 6(3a) = 14
a + 18a = 14
19a = 14
a = 14/19
So, the first term 'a' is 14/19, and the common difference 'd' is (3/19)(14/19), which simplifies to 42/361.
To find the twentieth term, we can use the equation for the nth term of a linear sequence:
Tn = a + (n-1)d
Substituting the values of 'a' and 'd', and setting n = 20, we get:
T20 = (14/19) + (20-1)(42/361)
= 14/19 + 19(42/361)
= 14/19 + 798/361
= (14*361 + 798*19)/(19*361)
= (5054 + 15162)/6859
= 20216/6859
= 20216/6859
So, the twentieth term of the linear sequence is 20216/6859.
To find the twentieth term of the linear sequence, we first need to determine the common difference (d) of the linear sequence.
Since we are given that the first, third, and ninth terms of the linear sequence are the first three terms of a geometric progression (G.P), let's denote these terms as a, b, and c.
In a linear sequence, the difference between consecutive terms is constant. Therefore, we can calculate the common difference (d) by finding the difference between any two consecutive terms of the linear sequence.
Let's calculate the common difference using the given information.
We know that the seventh term of the linear sequence is 14, which means it can be expressed as: a + 6d = 14.
Now, let's substitute the values of a, b, and c from the given information:
a = first term = first term of the G.P
b = third term = second term of the G.P
c = ninth term = third term of the G.P
This implies that a = a, b = ar, and c = ar^2, where r is the common ratio of the G.P.
Since a = a, b = ar, and c = ar^2, we have the following equations:
a + 2d = a,
a + 6d = b,
a + 18d = c.
Simplifying these equations, we can find a and d:
a + 2d = a ⇒ 2d = 0 ⇒ d = 0.
a + 6d = b ⇒ a + 0 = b ⇒ a = b.
a + 18d = c ⇒ a + 0 = c ⇒ a = c.
Therefore, we have determined that the common difference (d) of the linear sequence is 0. This means the sequence is not a linear sequence, but rather a constant sequence where each term is the same.
Knowing that all terms are the same (a = b = c), we can conclude that the twentieth term of the linear sequence will be equal to the first term, which is also equal to the seventh term and is equal to 14.
Hence, the twentieth term of the linear sequence is 14.