An arithmetic & geometric progression question.

In summary, the first term of the geometric series can be found by solving the equations derived from the given information, which can be written as a_q = (a_p + 10d) / (610 - 19d) / (20a_p + 10d).
  • #1
misogynisticfeminist
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I have an arithmetic series, with the sum of the first n terms to be 610. The 1st, 3rd and 11th terms of this AP is the same as the 3rd, 2nd and 1st term of a geometric series. Find the first term of the geometric series.

I have constructed 4 equations from this

[tex] a_p = a_q r^2 [/tex]

[tex]a_p+2d = a_q r [/tex]

[tex] a_p + 10d =a_q [/tex]

[tex] 20a_p +19d =610 [/tex]

where p represents the AP and Q represents the GP.

But i seem to have problems solving them simultaneously. Can anyone provide some insight?
 
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  • #2
Using the equations, we can solve for a_q: a_p = a_q r^2 a_p + 10d = a_q Substituting the second equation into the first gives us: a_p + 10d = a_q r^2 Rearranging this gives us: a_q = (a_p + 10d) / r^2 Now we can substitute this into the fourth equation: 20a_p + 19d = 610 Substituting the expression for a_q gives us: 20a_p + 19d = 610 20(a_p + 10d) / r^2 + 19d = 610 Solving for r^2 gives us: r^2 = (610 - 19d) / (20a_p + 10d) Substituting this into the expression for a_q gives us: a_q = (a_p + 10d) / (610 - 19d) / (20a_p + 10d) Therefore, the first term of the geometric series is: a_q = (a_p + 10d) / (610 - 19d) / (20a_p + 10d)
 
  • #3


It seems like you are on the right track with setting up the equations, but there may be a mistake in your second equation. Since the 1st, 3rd, and 11th terms of the arithmetic series are the same as the 3rd, 2nd, and 1st terms of the geometric series, the second equation should be:
a_p + 2d = a_q r^2

Then, you can solve the system of equations simultaneously using substitution or elimination.

Substituting the third equation into the fourth equation, we get:
20(a_q r - 2d) + 19d = 610

Simplifying, we get:
20a_q r - 41d = 610

Next, we can substitute the first equation into the second equation:
a_q r^2 + 2d = a_q r

Simplifying, we get:
a_q (r^2 - r) = -2d

We can then substitute this into the equation we got by substituting the third equation into the fourth equation:
20(-2d) - 41d = 610

Simplifying, we get:
-81d = 610

Solving for d, we get:
d = -610/81

Now, we can substitute this value for d into any of the previous equations to solve for a_q or r.

For example, substituting d into the third equation, we get:
a_q = (610/81) - (10/81)
a_q = 600/81

To find the first term of the geometric series, we can substitute these values into the first equation:
a_p = (600/81) * (600/81)^2
a_p = (600/81)^3
a_p = 2500/27

Therefore, the first term of the geometric series is 2500/27.
 

1. What is an arithmetic progression?

An arithmetic progression is a sequence of numbers in which each term is obtained by adding a fixed number to the previous term. For example, the sequence 3, 7, 11, 15, 19 is an arithmetic progression with a common difference of 4.

2. What is a geometric progression?

A geometric progression is a sequence of numbers in which each term is obtained by multiplying the previous term by a fixed number. For example, the sequence 2, 6, 18, 54, 162 is a geometric progression with a common ratio of 3.

3. How do you find the common difference or ratio in a progression?

To find the common difference in an arithmetic progression, subtract any term from the term before it. To find the common ratio in a geometric progression, divide any term by the term before it.

4. How do you calculate the sum of terms in a progression?

For an arithmetic progression, the sum of the first n terms can be calculated using the formula S = (n/2)(2a + (n-1)d), where a is the first term and d is the common difference. For a geometric progression, the sum of the first n terms can be calculated using the formula S = a((1-r^n)/(1-r)), where a is the first term and r is the common ratio.

5. How can progressions be applied in real life?

Progressions are commonly used in mathematics, engineering, and finance to model real-life situations. For example, they can be used to predict population growth, analyze compound interest, or design machinery with moving parts. Additionally, progressions are used in computer science to create algorithms and in music to create harmonies.

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