Question

In a G.P. of real numbers, it is given that $T_1+T_2+T_3+T_4=30$ and $T_1^2+T_2^2+T_3^2+T_4^2=340$. Determine the G.P.

• A. $2,4,8,16,32,...$
• B. $16,8,4,...$
• C. $1,2,4,...$
• D. $8,2,0.5,...$

Answer 1

Answer: (A), (B)

$2,4,8,16,32,...$
$16,8,4,...$

$T_1+T_2+T_3+T_4=30$

${T_1}^2+{T_2}^2+{T_3}^2+{T_4}^2=340$

$t_1=a,ratio=r$

$a(1+r+r^2+r^3)=30\Rightarrow \frac{a(r^4-1)}{r-1}=30-\left(1\right)$

$a^2(1+r^2+r^4+r^6)=340\Rightarrow \frac{a^2(r^8-1)}{r^2-1}=340-\left(2\right)$

by dividing $\frac{eq.2}{{\left(eq.1\right)}^2}$

$\Rightarrow \frac{\frac{a^2(r^8-1)}{r^2-1}}{\frac{a^2{(r^4-1)}^2}{{(r-1)}^2}}=\frac{340}{900}=\frac{17}{45}$

$\Rightarrow \frac{a^2(r^8-1)\times {(r-1)}^2}{(r^2-1)\times {(r^4-1)}^2}=\frac{r^4+1}{r^4-1}\times \frac{r-1}{r+1}=\frac{17}{45}$

$\Rightarrow \frac{r^4+1}{r^3+r^2+r+1}\times \frac{1}{r+1}=\frac{17}{45}$

$\Rightarrow \frac{\frac{r^4+1}{r^4}}{\frac{(r^3+r^2+r+1)}{r^3}\frac{r+1}{r}}=\frac{1+\frac{1}{r^4}}{(1+\frac{1}{r}+\frac{1}{r^2}+\frac{1}{r^3})(1+\frac{1}{r})}=\frac{17}{45}$

We can see clearly $2$ satisfy the equation

So $r=2$ and so as equation is symmetric in $r$ and $\frac{1}{r}$, $r$ can also be $\frac{1}{r}$

By $\left(1\right)$ if $r=2\frac{a(r^4-1)}{r-1}=30$

$\Rightarrow a(2^4-1)=30$

$\Rightarrow a=2$

So, $2,4,8,16,..$

if $r=\frac{1}{2}\frac{a({\left(\frac{1}{2}\right)}^4-1)}{\frac{1}{2}-1}=30$

$\frac{a\left(\frac{-15}{16}\right)}{-\frac{1}{2}}=30$

$\Rightarrow a=16$

hence we get $16,8,4,2,...$