Questions on geometric series

Question

1. Consider a three-term arithmetic sequence whose terms are $t_1, t_2, t_3$. The related sequence $t_1+8, t_2,t_3,$ is geometric and the sum of its terms is 26. Find the terms in the sequence

For the first question i am thinking this: the common ratio is $\frac{t_2}{t_1+8}$ which has to be equal to $\frac{t_3}{t_2}$

so $t_2^2= t_3(t_1+8)$

Not sure where to go from here or if i'm even correct.

The sum is 26 so $26 = \frac{ t_1+8 (1-\frac{t_3}{t_2})^3}{1-\frac{t_3}{t_2}}$

Answer 1

Your terms are $$ x-d,x,x+d$$ and you want $$ x-d+8, x,x+d$$ to form a geometric progression. That is $$ \frac {x}{x-d+8}=\frac {x+d}{x}$$

We get $$(x+d)(x-d+8)=x^2$$ Simplify to get $$d^2=8(x+d)$$

You have infinitely many solutions for example $d=12$ and $x=6$ imply $$ -6,6,18$$ for the arithmetic progression and $$2,6,18$$ for the geometric progression.

Answer 2

1. a)$t_2-d, t_2, t_2+d$

   b) $t_2-d+8, t_2, t_2+d$

so $3t_2+8=26$ so $t_2=6$

so $6^2=(14-d)(6+d)$

solving 2 solutions. $(t_1,t_2,t_3)=$ $(10,6,2)$ and $(-6,6,18)$

Answer 3

It is easier to eliminate $t2$ at first and then solve the quadratic equation

$$ 26-t2-t3= 2\, t2-t3 $$

when $t3$ automatically cancels out

$$t2= \dfrac{26}{3}$$

$$ t1+t3= \dfrac{52}{3},\quad (t1+8) \,t3= \dfrac{26}{3}$$

Eliminate either $t1$ or $t3$ to solve quadratic

$$ \ t1 =\dfrac{14\pm 16 \sqrt 3}{3},\quad t3= \dfrac{38\mp 16 \sqrt 3}{3}.$$