Question

How to write for the $n^{th}$ term of this geometric sequence? Then how to use that formula for ${T_n}$ to find ${T_8}$ (the eighth term of the sequence)? 4, -4, 4, -4…..

Answer 1

In these types of questions which the given series is in geometric series we will make use of formula of ${n^{th}}$term in the geometric series formula that can be can be written as, ${T_n} = a{r^{n - 1}}$, where is the first term and $r$ is the common ratio which is given by $\dfrac{{{a_{n + 1}}}}{{{a_n}}}$, now by substituting the values in the formulas we will get the required result.

Complete step by step answer:
The sequence is 4, -4, 4, -4…..
The given sequence is in geometric progression and its common ratio is given by $\dfrac{{{a_{n + 1}}}}{{{a_n}}}$ so here common ratio is -1.
So, we know that the ${n^{th}}$ term in geometric sequence is given by ${T_n} = a{r^{n - 1}}$,
Now we have to find the ${8^{th}}$ term of the given sequence, we know The ${n^{th}}$ term in geometric progression is given by ${T_n} = a{r^{n - 1}}$,,
By substituting the values in the formula we get, here $n= 8$, $a = 4$, $r = - 1$, we get,
$\Rightarrow {T_8} = \left( 4 \right){\left( { - 1} \right)^{8 - 1}}$,
Now simplifying we get,
$\Rightarrow {T_8} = \left( 4 \right){\left( { - 1} \right)^7}$,
Again simplifying we get,
$\Rightarrow {T_8} = \left( 4 \right)\left( { - 1} \right)$,
Now multiplying we get,
$\Rightarrow {T_8} = - 4$
So, the ${8^{th}}$ term is $- 4$.

$\therefore$ The ${8^{th}}$ term in geometric sequence is given by ${T_n} = a{r^{n - 1}}$ and the ${8^{th}}$ term of the sequence 4, -4, 4, -4….. will be equal to $- 4$.

Note: As there are 3 types of series i.e., Arithmetic series, Geometric series and Harmonic series, students should not get confused which series to be used in what type of questions, as there are many formulas related to each of the series, here are some useful formulas related to the above series,
Sum of the $n$ terms in A. P is given by,${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$, where $n$ is common difference, $a$ is the first term.
The ${n^{th}}$ term In A.P is given by ${T_n} = a + \left( {n - 1} \right)d$,
Sum of the $n$ terms in GP is given by, ${S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}$, where $r$ is common ratio, $a$ is the first term.
The ${n^{th}}$ term In G.P is given by ${T_n} = a{r^{n - 1}}$,
If a, b, c are in HP, then b is the harmonic mean between a and c.
In this case, $b = \dfrac{{2ac}}{{a + c}}$.