# Thread: Geometric Progression- Ans to confirm only

1. ## Geometric Progression- Ans to confirm only

Question:
if 1+y, 1+3y, 1+4y (y cant be 0..sorry i dont know how to write it in math equation here ) are three consecutive terms of a Geometric Progression. Find
(a) the value of y
(b) the sum to infinity.

the ans from book
(a) -1/5
(b) 8/5

but i have calculate and get this ans
(a) 1/5
(b) -18/5

2. Originally Posted by nikk
Question:
if 1+y, 1+3y, 1+4y (y cant be 0..sorry i dont know how to write it in math equation here ) are three consecutive terms of a Geometric Progression. Find
(a) the value of y
(b) the sum to infinity.

the ans from book
(a) -1/5
(b) 8/5

but i have calculate and get this ans
(a) 1/5
(b) -18/5

In a geometric progression: $\frac{t_{n + 1}}{t_n} = r$.

So $\frac{1 + 3y}{1 + y} = r$ and $\frac{1 + 4y}{1 + 3y} = r$.

Therefore $\frac{1 + 3y}{1 + y} = \frac{1 + 4y}{1 + 3y}$

$(1 + 3y)^2 = (1 + y)(1 + 4y)$

$1 + 6y + 9y^2 = 1 + 5y + 4y^2$

$5y^2 + y = 0$

$y(5y + 1) = 0$

$y = 0$ or $5y + 1 = 0$

$y = 0$ or $y = -\frac{1}{5}$.

Since $y \neq 0$ that means $y = -\frac{1}{5}$.

Now to find the sum to infinity, we need $a$ and $r$

$r = \frac{1 + 3y}{1 + y}$

$= \frac{1 + 3\left(-\frac{1}{5}\right)}{1 - \frac{1}{5}}$

$= \frac{1 - \frac{3}{5}}{\frac{4}{5}}$

$= \frac{\frac{2}{5}}{\frac{4}{5}}$

$= \frac{1}{2}$.

Are we assuming that $a = 1 + y$?

If so $a = \frac{4}{5}$.

Now, since $|r| < 1$ we can work out the sum to infinity...

$S_{\infty} = \frac{a}{1 - r}$

$= \frac{\frac{4}{5}}{1 - \frac{1}{2}}$

$= \frac{\frac{4}{5}}{\frac{1}{2}}$

$= \frac{8}{5}$.

3. tq for the post ..now i can see where my calculation wrong

can u guide me how to make this equation in latex look more bigger

http://www.mathhelpforum.com/math-he...ze-bigger.html