Question

the product of three terms of a G.P is 512 if 8 is added to the first term and 6 to the second term find the new term form an A.P

Answer 1

The product of three terms of a GP is 512.
Suppose the terms of the GP are: $\frac{a}{r}$, $a$ and $ar$


Then, we have:

$Product = 512 \\ \\ \implies \frac{a}{r} \times a \times ar = 512 \\ \\ \implies a^3 = 512 \\ \\ \implies a = \sqrt[3]{512} \\ \\ \implies \boxed{\bold{a = 8}}$


Now, the terms become:

$\frac{8}{r}$, $8$ and $8r$


Now, we are given that when 8 is added to the first term and 6 is added to the second term, the resulting terms are in AP.


This means that:

$\frac{8}{r}+8$, $8+6$, $8r$ are in AP

$\implies \frac{8}{r}+8$, $14$, $8r$ are in AP.


$\displaystyle \implies 14 = \frac{\left( \frac{8}{r}+8 \right) + \left( 8r \right)}{2} \\ \\ \\ \implies 14 = \frac{4}{r} + 4 + 4r \\ \\ \\ \implies \frac{4}{r} + 4r = 10 \\ \\ \\ \implies 4\left( \frac{1}{r}+r \right) = 10 \\ \\ \\ \implies \frac{1}{r}+r = \frac{10}{4} \\ \\ \\ \implies \frac{1+r^2}{r} = \frac{5}{2} \\ \\ \\ \implies 2 + 2r^2 = 5r \\ \\ \\ \implies 2r^2-5r+2=0 \\ \\ \\ \implies 2r^2-4r-r+2=0 \\ \\ \\ \implies 2r(r-2)-1(r-2)=0 \\ \\ \\ \implies (2r-1)(r-2)=0$


$\implies r-2=0 \quad OR \quad 2r-1=0 \\ \\ \\ \implies \boxed{r=2} \quad OR \quad \boxed{r=\frac{1}{2}}$



If $r=2$, then the terms of GP become:

$\rightarrow \frac{8}{r} = \frac{8}{2} = 4 \\ \\ \rightarrow 8 \\ \\ \rightarrow 8r = 8\times 2 = 16 \\ \\ \\ \implies \boxed{\textbf{GP is 4, 8, 16}} \\ \\ \\ \implies \boxed{\textbf{AP is 12, 14, 16}}$


If $r=\frac{1}{2}$, then the terms of GP become:

[tex]\rightarrow \frac{8}{r} = \frac{8}{\frac{1}{2}} = 16 \\ \\ \rightarrow 8 \\ \\ \rightarrow 8r = 8\times \frac{1}{2} = 4 \\ \\ \\ \implies \boxed{\textbf{GP is 16, 8, 4}} \\ \\ \\ \implies \boxed{\textbf{AP is 24, 14, 4}}[/tex]

Answer 2

Step-by-step explanation:

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