Question

The ${{1025}^{\text{th}}}$ term of the sequence 1, 2, 2, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8, … is ${a^{10}}$ . Find a.

Answer 1

The given sequence is formed by repeating n terms of a G.P. 1, 2, 4, 8, … n times.
So, here $a = 1$ and $r = 2$.
Then, find the sum of n terms using the formula ${S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$.
Also, ${S_n} \geqslant 1025$, and find n.
Thus, we will get the required answer.

Complete step by step solution:
Let m be the value of the number of times ${a^{10}}$ is repeated in the given sequence.
$\therefore$m = 1, 2, 4, 8, … = ${2^0}$ , ${2^1}$ , ${2^2}$ , ${2^4}$ , …
It is clear that sequence m is a G.P, in which $a = 1$, $r = 2$.
So, the ${{n}^{\text{th}}}$ term of G.P. will be ${2^{n - 1}}$ , where n = 1, 2, 3, …
Now, sum of n terms of a G.P. is given by ${S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$
Substituting values of a and r in ${S_n}$
$\therefore {S_n} = 1\left( {\dfrac{{{2^n} - 1}}{{2 - 1}}} \right)$
$\therefore {S_n} = {2^n} - 1$
The ${{1025}^{\text{th}}}$ term of the sequence will be minimum m such that ${S_n} \geqslant 1025$.
${S_n} \geqslant 1025$
$\therefore {2^n} - 1 \geqslant 1025 \\\ \therefore {2^n} \geqslant 1025 + 1 \\\ \therefore {2^n} \geqslant 1026 \\\$
$\therefore$ The minimum value of m for which ${2^n} \geqslant 1026$ is 2048 $= {2^{11}}$.
Now, ${2^{11}}$ is the ${{12}^{\text{th}}}$ term of the given G.P., because the nth term of G.P. is ${2^{n - 1}}$.
$\therefore$The ${{1025}^{\text{th}}}$ term of sequence is the ${{11}^{\text{th}}}$ term of the G.P. which is ${2^{10}}$ .

Now, comparing ${2^{10}}$ with ${a^{10}}$, we get $a = 2$.

Note:
The given sequence is the type of sequence in which n consecutive terms have the value n. The terms of the given sequence are in a G.P. 1, 2, 4, 8, …, in which 1 is repeated 1 time, 2 is repeated 2 times, 4 is repeated 4 times, 8 is repeated 8 times and it continues so on. Also, the value of the term is the first position of that term i.e. 1 is at ${{1}^{\text{st}}}$ position, 2 is at ${{2}^{\text{nd}}}$ position, 4 is at ${{4}^{\text{th}}}$ position, 8 is at ${{8}^{\text{th}}}$ position, … and n is at ${{n}^{\text{th}}}$ position.