Arithmetic series sum 610, three terms form a geometric series

[misogynisticfeminist]

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

$$a_p = a_q r^2$$

$$a_p+2d = a_q r$$

$$a_p + 10d =a_q$$

$$20a_p +19d =610$$

where p represents the AP and Q represents the GP.

But i seem to have problems solving them simultaneously. Can anyone provide some insight?

 

[member 553083]

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)

 

[thepatientmental]

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.