The sequence 8, a, b, 36 is arithmetic with respect to the first 3 terms and geometric to last 3 - what can a and b be?

Question

The sequence $8, a, b, 36$ is an arithmetic sequence with respect to the first 3 terms and geometric with respect to the last 3 terms. What are all possible values of a and b? My attempts at solving this problem have so far failed. Please help. Thank you very much.

Answer 1

I have form the first condition: $$a=8+d$$ and $$b=8+2d$$ and from the second $$b=aq$$ $$36=aq^2$$ You get the equations $$b=8+2(a-8)$$ and $$36=a\cdot \left(\frac{b}{a}\right)^2$$

Answer 2

Hint: When is a sequence $x,y,z$:

-arithmetic?

-geometric?

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You should get $2a=8+b$ and $b^2=36a$. Now solve the system...

Answer 3

The first $3$ terms of the sequence $8, a, b, 36$ are in AP

So $ a-8 = b-a$ as the difference is same.

So you have $2a= b +8$ ...(1)

Similarly, the last $3$ terms are in GP,

So, the ratio of third to second term will be equal to the ratio of fourth to third term

$\therefore \dfrac{b}{a} = \dfrac{36}{b} \implies b^2 = 36a $

Solve them, and you should get $b = 24$ or $b=-6$ put these values in (1) to get corresponding values of a