The fifth ninth and sixteenth terms of a linear sequence ( ap) are consecutive terms of an exponential sequence

1 find the common difference of the linear sequence in terms of the 1st term
2 show that the 21st 37th and 65th terms of the linear sequence are consecutive terms of an exponential sequence whose common ratio is 7/4

(a+8d)/(a+4d) = (a+15d)/(a+8d)

so, 3a=4d

#1. a = 4/3 d

#2. You can pick your values for a and d, say, a=4, d=3
So now form the desired terms, and show that they have a common ratio.

Ap a

Gp B
A +4d=b
A+8d=br
A+15d =br ^2
A=4/3d,b=49,r=7÷4
Picking value for d=3
A=3,b=12
Terms of g.p=12,21,147÷7

6:5

Ap a

Gp B
A +4d=b
A+8d=br
A+15d =br ^2
A=4/3d,b=49,r=7÷4
Picking value for d=3
A=3,b=12
Terms of g.p=12,21,147÷7

1. The common difference of a linear sequence can be found by subtracting the first term from the second term. Let's say the first term of the linear sequence is "a," and the common difference is "d." The second term would be "a + d."

Since the fifth, ninth, and sixteenth terms of the linear sequence are consecutive terms of an exponential sequence, we can equate them:

5th term: a + 4d
9th term: a + 8d
16th term: a + 15d

For consecutive terms of an exponential sequence, we know that each term is multiplied by a common ratio. So, we can set up the equation:

(a + 8d) / (a + 4d) = (a + 15d) / (a + 8d)

Cross-multiplying and simplifying, we get:

(a + 8d)(a + 8d) = (a + 4d)(a + 15d)

a^2 + 16ad + 64d^2 = a^2 + 19ad + 60d^2

16ad - 19ad = 60d^2 - 64d^2

-3ad = -4d^2

Dividing both sides by "d" (assuming "d" is not zero), we get:

-3a = -4d

Therefore, the common difference of the linear sequence in terms of the first term is:
d = (3/4)a

2. To show that the 21st, 37th, and 65th terms of the linear sequence are consecutive terms of an exponential sequence with a common ratio of 7/4, we can use a similar approach.

Let's say the 21st term of the linear sequence is "a + 20d," the 37th term is "a + 36d," and the 65th term is "a + 64d."

Using the equation for consecutive terms of an exponential sequence, we have:

(a + 36d) / (a + 20d) = (a + 64d) / (a + 36d)

Cross-multiplying and simplifying, we get:

(a + 36d)(a + 36d) = (a + 20d)(a + 64d)

a^2 + 72ad + 1296d^2 = a^2 + 84ad + 1280d^2

72ad - 84ad = 1280d^2 - 1296d^2

-12ad = -16d^2

Dividing both sides by "d" (assuming "d" is not zero), we get:

-12a = -16d

Therefore, the common difference of the linear sequence in terms of the first term is:
d = (3/4)a

Hence, we have shown that the 21st, 37th, and 65th terms of the linear sequence are consecutive terms of an exponential sequence with a common ratio of 7/4.

To find the common difference of the linear sequence in terms of the first term (a), we can use the formula for the nth term of a linear sequence.

1. The formula for the nth term of a linear sequence is given by:
an = a + (n - 1)d

Here, a represents the first term of the sequence, n represents the position of the term, and d represents the common difference.

Given that the fifth, ninth, and sixteenth terms of the linear sequence are consecutive terms of an exponential sequence, we can set up the following equations:

a + (5 - 1)d = a + (9 - 1)d = a + (16 - 1)d

Simplifying each equation:

1a + 4d = 1a + 8d = 1a + 15d

From the first two equations, we can subtract the left-hand side of the second equation from the left-hand side of the first equation, and the right-hand side of the second equation from the right-hand side of the first equation:

1a + 4d - (1a + 8d) = 0
-4d = -4d

This implies that d cancels out, meaning that the common difference (d) is eliminated from the equation. Therefore, the common difference of the linear sequence, in terms of the first term (a), is 0.

Now, let's move on to the second part of the question.

2. To show that the 21st, 37th, and 65th terms of the linear sequence are consecutive terms of an exponential sequence with a common ratio of 7/4, we can apply similar reasoning.

We'll use the formula for the nth term of an exponential sequence:

an = a * r^(n-1)

Here, a represents the first term, n represents the position of the term, and r represents the common ratio.

Using the given information, we can set up the following equations:

a = a * (7/4)^(21-1)
a = a * (7/4)^(37-1)
a = a * (7/4)^(65-1)

Simplifying each equation:

a * (7/4)^20 = a * (7/4)^36 = a * (7/4)^64

We can again subtract the left-hand side of the second equation from the left-hand side of the first equation, and the right-hand side of the second equation from the right-hand side of the first equation:

a * (7/4)^20 - a * (7/4)^36 = 0
a * (7/4)^36 - a * (7/4)^64 = 0

We can factor out 'a' from each equation:

a * [(7/4)^20 - (7/4)^36] = 0
a * [(7/4)^36 - (7/4)^64] = 0

For these equations to hold true, either 'a' must be equal to zero or the expressions in brackets must be equal to zero. However, since 'a' cannot be zero (as it is the first term of the sequence), the equations cannot be satisfied.

Therefore, it is not possible to show that the 21st, 37th, and 65th terms of the linear sequence are consecutive terms of an exponential sequence with a common ratio of 7/4.