Let a_1,a_2,a_3, ... ... . be a G.P. of increasingn positive numbers. Let the sum of its 6^th andn 8^th terms be 2 and the product of itsn 3^rd and 5^th terms ben 1/9. Thenn 6( a_2 + a_4 )( a_4 + a_6 ) is equaln to

n n a r^5+a r^7=2 n (a r^2)(a r^4)=(1 / 9) n a^2 r^6=(1 / 9)n nn nNow, r>0nnar^5(1+ r^2)=2nn nNow, ar^3=(1 / 3) or -(1 / 3)( rejected )nn n r^2=2 n r=sqrt(2) n a=frac16 sqrt(2)n nn nNow, 6(a_2+a_4)(a_4+a_6)nn n 6(a r+a r^3)(a r^3+a r^5) n 6 a^2 r^4(1+r^2) n 6((1 / 36.2))(4)(9)=3n n

Let a_{1},a_{2},a_{3},. be a G.P. of increasing posi... | Sequence & Series (Mathematics)

\n\begin{aligned}\n& a r^5+a r^7=2 \\\n& \left(a r^2\right)\left(a r^4\right)=\frac{1}{9} \\\n& a^2 r^6=\frac{1}{9}\n\end{aligned}\n

\n \nNow,

\mathrm{r}>0

\n\operatorname{ar}^5\left(1+\mathrm{r}^2\right)=2\n

\n \nNow,

\mathrm{ar}^3=\frac{1}{3}

-\frac{1}{3}(

rejected

)

\n\begin{aligned}\n& r^2=2 \\\n& r=\sqrt{2} \\\n& a=\frac{1}{6 \sqrt{2}}\n\end{aligned}\n

\n \nNow,

6\left(a_2+a_4\right)\left(a_4+a_6\right)

\n\begin{aligned}\n& 6\left(a r+a r^3\right)\left(a r^3+a r^5\right) \\\n& 6 a^2 r^4\left(1+r^2\right) \\\n& 6\left(\frac{1}{36.2}\right)(4)(9)=3\n\end{aligned}\n

Explanation

Let a_{1},a_{2},,a_{n} be a given A.P. whose common difference is an integer and S_{n} = a_{1} + a_{2} + + a_{n}. If a_{1} = 1,a_{n} = 300 and 15 \leq n \leq 50, then the ordered pair ( S_{n - 4},a_{n - 4} ) is equal to

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Let 3,a,b,c be in A.P. and 3,a - 1,b + 1,c + 9 be in G.P. Then, the arithmetic mean of a,b and c is

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