Help with sequence: a_1 + a_3 = 280, a_4 + a_6 = 35

[johnk]

I need help solving a geometric sequence which is defined by:

first term + third term = 280
fourth term + sixth term = 35

I need to find the ratio of the sequence and the first term, this problem is solved in the book in which it appears, but the solution says only:

q = ratio, a = first term
\(\displaystyle \large a + aq^2 = 280\\
aq^3 + aq^5 = 35\)
and thus
$\displaystyle \large q^3 = \frac{1}{8};\ \ q = \frac{1}{2}$

What's the intermediate step?

What I tried was:
(from the first equation) $\displaystyle \large q = \sqrt{\frac{280}{a}-1}$
and then substituting that for q in the second equation, but that gets very complex and I get stuck, it seems there is an easier way.

 

[galactus]

The nth term of a geometric sequence is given by:

\(\displaystyle \L\\a_{n}=a_{1}r^{n-1}\)

first and third terms equal 280:

\(\displaystyle \L\\a_{1}+a_{1}r^{2}=280\)

The fourth and sixth equal 35:

\(\displaystyle \L\\a_{1}r^{3}+a_{1}r^{5}=35\)

So, you have the system:

\(\displaystyle \L\\a_{1}(1+r^{2})=280\)
\(\displaystyle \L\\a_{1}r^{3}(1+r^{2})=35\)

Solve for $\displaystyle a_{1} \;\ and \;\ r$

 

[johnk]

Thank you for your reply.

I hope I understood you correctly, I did this:

$\displaystyle \Large a_1 = \frac{280}{1+r^2}$ (from the first equation)

$\displaystyle \Large \frac{280}{1+r^2}\cdot r^3 \cdot (1+r^2) = 35$ (substituted to the second equation)

\(\displaystyle \Large 280r^3 \cdot (1+r^2) = 35 + 35r^2\\
280r^5 + 280r^3 - 35r^2 - 35 = 0\\
56r^5 + 56r^3 - 7r^2 - 7 = 0\\
7r^2\cdot (8r^3+8r-1)-7 = 0\)

What do I do with that? Or am I completely on the wrong track? English is not my first language, I'm sorry if I misunderstood, in any case please give me some more help.

 

[galactus]

You're making this more difficult than need be.


\(\displaystyle \L\\1+r^{2}=\frac{280}{a_{1}}\)

Sub into the other equation:

\(\displaystyle \L\\a_{1}r^{3}(\frac{280}{a_{1}})=35\)

\(\displaystyle \L\\r=\sqrt[3]{\frac{35}{280}}=\frac{1}{2}\)

Now, you have r, can you find a1?.

 

[johnk]

Ahh, I never thought about substituting whole (1+r^2)... That's a nice trick . I always substituted just a1 or r. It's easy now knowing that. a1 = 224

Thank you very much

 

[galactus]

There ya' go. You got it. The sequence should be easy now.

 

[Denis]

\(\displaystyle \L\\a_{1}(1+r^{2})=280\)
-----------------------------
\(\displaystyle \L\\a_{1}r^{3}(1+r^{2})=35\)

Still shorter here is a division; a1(1 + r^2) cancels out, 280/35 = 8:
1 / r^3 = 8
r^3 = 1/8
r = 1/2

 

[soroban]

Hello, John!

With a geometric sequence or series,
. . there is a "trick" that can often be used.

Geometric sequence:
First term + third term = 280
Fourth term + sixth term = 35
Find the ratio and the first term.

We are given: \(\displaystyle \L\:\begin{array}{ccccccccc}a\,+\,ar^2 & = & 280 &\;\Rightarrow\; & a(1\,+\,r^2) & = & 280 & \;[1] \\
ar^3\,+\,ar^5 & = & 35 & \;\Rightarrow\; & ar^3(1\,+\,r^2) & = & 35 & \;[2] \end{array}\)


Divide [2] by [1]: \(\displaystyle \L\:\frac{ar^3(1\,+\,r^2)}{a(1\,+\,r^2)} \:=\:\frac{35}{260}\;\;\Rightarrow\;\;r^3\:=\:\frac{1}{8}\)

. . Therefore: \(\displaystyle \L\:\fbox{r \:=\:\frac{1}{2}}\)

Substitute into [1]: \(\displaystyle \L\:a\left(1\,+\,\frac{1}{4}\right)\:=\:280\;\;\Rightarrow\;\;\frac{5}{4}a\:=\:280\)

. . Therefore: \(\displaystyle \L\:\fbox{a\:=\:224}\)

Edit: Too fast for me, Denis!