If a_n = (3 / 4) - ( (3 / 4) )^2 + ( (3 / 4) )^3 + ... ( - 1)^n - 1( (3 / 4) )^nn and b_n = 1 - a_n, then find the least natural numbern n_0 such that b_n > a_nforall n >= n_0

nna_n=((3 / 4))-((3 / 4))^2+((3 / 4))^3+ * s .+(-1)^n-1((3 / 4))^nnnnnG.P. with common ratio -(1 / 3)nnnn nna_n=frac(3 / 4)[1-(-(3 / 4))^n]1-(-(3 / 4))=(3 / 7)[1-(-(3 / 4))^n] nnb_n>a_n => 1-a_n>a_n => a_n

If a_{n} = \frac{3}{4} - ( \frac{3}{4} )^{2} + ( \frac{... | Sequence & Series (Mathematics)

\n\na_n=\left(\frac{3}{4}\right)-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3+\cdots .+(-1)^{n-1}\left(\frac{3}{4}\right)^n\n\n

\n\nG.P. with common ratio

-\frac{1}{3}

\n\n

\n\n\begin{gathered}\n\na_n=\frac{\frac{3}{4}\left[1-\left(-\frac{3}{4}\right)^n\right]}{1-\left(-\frac{3}{4}\right)}=\frac{3}{7}\left[1-\left(-\frac{3}{4}\right)^n\right] \\\n\nb_n>a_n \Rightarrow 1-a_n>a_n \Rightarrow a_n<\frac{1}{2}\n\n\end{gathered}\n\n

\n \nFrom (1) and (2),

(-3)^{n+1}<2^{2 n-1}

\n\nIf

n

is even L.H.S. is negative, R.H.S. is positive and the inequality certainly holds.\n\nIf

n

is odd, the above holds for

n \geq 7

\n\n∴ The least natural number

n_0

for which it holds is 6 .\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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