Finding a Sequence of Number

[stevecowall]

Question: The terms y1, y2, y3, y4, y5 form an arithmetic sequence whose sum is 920. The terms a2, a3, a5 form a geometric sequence. Find the sum of all possible values of a5.

Problems have been solved by soroban

 

[mmm4444bot]

The sum of a1, a2, a3, a5 is 920.

a + x + a + 2x + a + 3x + a +4x + a + 4x + a + 5x = 920

Typographical errors, here.

It appears that the element a4 moved from the first line into the second.



184/(184 - x) = (184 + 2x)/184

Solve for x so I got x = 0, -92

Your solution for x contains a sign error. Fix that, and it should all work out.

Great set-up, BTW.

 

[stevecowall]

Already fixed but i still cant figure out the correct answer

 

[soroban]

Hello, stevecowall!

The terms a1, a2, a3, a4, a5 form an arithmetic sequence whose sum is 920.
The terms a2, a3, a5 form a geometric sequence.
Find the sum of all possible values of a5.

Let \(\displaystyle a\) = first term, \(\displaystyle d\) = common difference.

The arithmetic sequence is: .\(\displaystyle a,\:a+d,\:a+2d,\:a+3d,\:a+4d\)

Its sum is 920: .\(\displaystyle 5a + 10d \,=\,920 \quad\Rightarrow\quad a + 2d \,=\,184\) .[1]


\(\displaystyle a_2,\,a_3,\,a_5\) form a geometric sequence.

Hence: .\(\displaystyle \dfrac{a_2}{a_3} = \dfrac{a_3}{a_5} \quad\Rightarrow\quad \dfrac{a+d}{a+2d} \,=\,\dfrac{a+2d}{a+4d}\)

. . . . . . \(\displaystyle (a+d)(a+4d) \:=\a+2d)(a+2d)\)

. . . . . . \(\displaystyle a^2 + 5ad + 4d^2 \:=\:a^2 + 4ad + d^2\)

. . . . . . . . . . . . . . . \(\displaystyle ad \:=\:0\)

Hence: .\(\displaystyle a = 0\,\text{ or }\,d = 0\)


If \(\displaystyle a = 0\), substitute into [1]:..\(\displaystyle d = 92\)

The sequence is: .\(\displaystyle 0,\:92,\;184,\;276,\;\boxed{368}\)


If \(\displaystyle d = 0\), substitute into [1]: .\(\displaystyle a = 184\)

The sequence is: .\(\displaystyle 184,\;184,\;184,\;184,\;\boxed{184}\)


The sum of all possible values of \(\displaystyle a_5\) is: .\(\displaystyle 368 + 184 \,=\,552\)

 

[stevecowall]

Thanks soroban and mmm4444bot,

This community is so awesome because of you guys ... Thank you so much