The first, third and ninth terms of a linear sequence are the first three terms of an exponential sequence. If the seventh term of the linear sequence is 14 calculate the twentieth term of the linear sequence and the sum of the first twelve terms of the exponential sequence

since GP's have a common ratio,

r = (a+2d)/a = (a+8d)/(a+2d)
a+6d = 14
So, solve for a and d, and then find
a+19d
S12 = a(r^12 - 1)/(r-1)

The 7th term of an AP is 4 and its Common difference is -4. What is its first term

1 point
16
20
24
28

I want this question solved in fully understandable, but where they wrote 14 is 4 so the 14 is error

To be solved

To solve this problem, let's first find the common difference of the linear sequence.

Given that the seventh term of the linear sequence is 14, we can use the formula for the nth term of a linear 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.

We know that the seventh term (T7) is 14, and the first term (a) is the third term of the exponential sequence. So, substituting these values into the formula:

14 = a + (7 - 1)d
14 = a + 6d ---(1)

Now, let's find the first term (a) and the common ratio (r) of the exponential sequence.

Given that the first, third, and ninth terms of the linear sequence are the first three terms of the exponential sequence, we have:

First term of exponential sequence, a_exp = a
Second term of exponential sequence, a_exp * r = T3
Third term of exponential sequence, a_exp * r^2 = T9

Dividing the second equation by the first equation, we get:

(a_exp * r) / a_exp = T3 / a
r = T3 / a ---(2)

Dividing the third equation by the second equation, we get:

(a_exp * r^2) / (a_exp * r) = T9 / T3
r = T9 / T3 ---(3)

Since both expressions on the right side of equations (2) and (3) are equal to r, we can set them equal to each other:

T3 / a = T9 / T3

Cross-multiplying:

T3 * T3 = a * T9
T3^2 = a * T9

Using equation (1), we can express a in terms of T3 and d:

14 = a + 6d
a = 14 - 6d

Substituting this expression for a in terms of T3, d, and T9:

T3^2 = (14 - 6d) * T9

Now we have two equations:
Equation (1): 14 = a + 6d
Equation (4): T3^2 = (14 - 6d) * T9

We can solve these equations for d and T3.

To find the common difference (d), we can solve equation (1) for d:

14 = (14 - 6d) + 6d
14 = 14
As the equation simplifies to 14 = 14, it means any value of d can work. So, we don't have a unique value for d.

To find T3, we can substitute the value of d back into equation (1):

14 = a + 6d
14 = a + 6(0)
14 = a
a = 14

Now that we have a = 14 and d could be any value, we can find the first few terms of the linear sequence.

The first term is a = 14.
The second term is a + d = 14 + d.
The third term is a + 2d = 14 + 2d.
The fourth term is a + 3d = 14 + 3d.

Continuing this pattern, the 20th term of the linear sequence will be:
a + 19d = 14 + 19d.

As for the exponential sequence, we know that the first twelve terms are the same as the first twelve terms of the linear sequence.
So, we can find the sum of the first twelve terms of the exponential sequence by adding up the values of the first twelve terms of the linear sequence.

The sum of the first twelve terms of the exponential sequence is:
(14 + 14 + d) + (14 + 2d) + (14 + 3d) + ... + (14 + 12d).

To find this sum, we can use the formula for the sum of an arithmetic series:

Sn = (n/2)(2a + (n - 1)d),

where Sn is the sum of the first n terms, a is the first term, d is the common difference, and n is the number of terms.

In this case, n = 12, a = 14, and d could be any value.

Substituting these values into the formula:

S12 = (12/2)(2 * 14 + (12 - 1)d)
S12 = 6(28 + 11d)
S12 = 168 + 66d

Therefore, the sum of the first twelve terms of the exponential sequence is 168 + 66d.

Finally, the 20th term of the linear sequence is given by:
a + 19d = 14 + 19d.

And the sum of the first twelve terms of the exponential sequence is:
168 + 66d.