Question

Consider the sequence $1,2,2,4,4,4,4,8,8,8,8,8,8,8,8,......$ and so on. Then ${{1025}^{th}}$ term will be
A. ${{2}^{9}}$
B. ${{2}^{11}}$
C. ${{2}^{10}}$
D. ${{2}^{12}}$

Answer 1

Hint: We first find the relation between the terms and its number of repetitions. We also find the starting position of a new number and its repletion number. This gives the span of the terms in which 1025 lies. We find the term from that.

Complete step by step solution:
We need to carefully look at the sequence $1,2,2,4,4,4,4,8,8,8,8,8,8,8,8,......$.
We can see not only the increasing digits are in G.P. form, but also the number of terms for a particular digit is also in G.P and being equal to the value of the digit itself.
1 is 1 time, 2 is 2 times, 4 is 4 times, 8 is 8 times and so on.
The individual terms are in the form ${{2}^{n}},n=0\left( 1 \right)...$
We need to find the position of a particular digit at the starting point.
We can see ${{2}^{r}}$ starts after the number of terms for the ${{2}^{0}}$ to ${{2}^{r-1}}$ is preceding it.
So, the number of terms in the span of ${{2}^{0}}$ to end of ${{2}^{r-1}}$ will be ${{2}^{0}}+{{2}^{1}}+...+{{2}^{r-1}}$.
The value of the common ratio is 2 for which the sum of the first the terms of the G.P. will be ${{S}_{n}}={{t}_{1}}\dfrac{{{r}^{n}}-1}{r-1}$. So, ${{2}^{0}}+{{2}^{1}}+...+{{2}^{r-1}}=1\times \dfrac{{{2}^{r}}-1}{2-1}={{2}^{r}}-1$
So, the new digit ${{2}^{r}}$ starts at the position of ${{2}^{r}}-1+1={{2}^{r}}$ and goes on for ${{2}^{r}}$ terms.
The nearest form like ${{2}^{r}}$ of 1025 is $1024={{2}^{10}}$.
So, the new digit ${{2}^{10}}$ starts at the position of ${{2}^{10}}=1024$ and goes on for 1024 times.
Therefore, ${{1025}^{th}}$ term will be ${{2}^{10}}$. The correct option is (C).

Note:
We can’t mix the starting position and the number of terms preceding it. Both are needed to find the starting number and the digit of a particular span.