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# If ${t_n} = 5 - 2n$, then ${t_{n - 1}} =$ (a) $2n - 1$  (b) $7 + 2n$  (c) $4 - 2n$  (d) $7 - 2n$

Last updated date: 17th Jun 2024
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Hint: Here, we need to find the value of ${t_{n - 1}}$. The equation ${t_n} = 5 - 2n$ is true for all values of $n$. We will substitute the required value of $n$ to obtain an algebraic expression and then simplify the expression to get the required answer.

We will substitute the value of $n$ in the given expression to get the required answer.
It is given that ${t_n} = 5 - 2n$.
This is true for all values of $n$.
For example, we can find the values of ${t_1}$, ${t_2}$, ${t_3}$, etc.
Substituting $n = 1$, we get
${t_1} = 5 - 2\left( 1 \right) = 5 - 2 = 3$
Similarly, we can find other values like
${t_2} = 5 - 2\left( 2 \right) = 5 - 4 = 1$
${t_3} = 5 - 2\left( 3 \right) = 5 - 6 = - 1$
Now, we need to find the value of ${t_{n - 1}}$.
We will substitute $n - 1$ for $n$ in the expression .
Substituting $n - 1$ for $n$, we get
${t_{n - 1}} = 5 - 2\left( {n - 1} \right)$
Multiplying the terms in the expression, we get
$\Rightarrow {t_{n - 1}} = 5 - 2n + 2$
Adding 5 and 2, we get
$\Rightarrow {t_{n - 1}} = 7 - 2n$
So, we get the value of ${t_{n - 1}}$ as $7 - 2n$.
The correct option is option (d).

Note: We can also solve this problem using the formula for ${n^{th}}$ term of an Arithmetic Progression.
We can find the values of ${t_1}$, ${t_2}$, ${t_3}$, etc.
Substituting $n = 1$, we get
${t_1} = 5 - 2\left( 1 \right) = 5 - 2 = 3$
Similarly, we can find other values like
${t_2} = 5 - 2\left( 2 \right) = 5 - 4 = 1$
${t_3} = 5 - 2\left( 3 \right) = 5 - 6 = - 1$
Now, this forms a sequence $3,1, - 1, \ldots \ldots \ldots ,\left( {5 - 2n} \right)$.
This forms an arithmetic progression with first term $a = 3$, common difference $d = 1 - 3 = - 2$, and number of terms $n$.
We know that ${n^{th}}$ term of an A.P. is given by the formula
${a_n} = a + \left( {n - 1} \right)d$, where $a$ is the first term, $d$ is the common difference, and $n$ is the number of terms.
Substituting $n - 1$ for $n$, we get, we get the ${\left( {n - 1} \right)^{th}}$ term of the A.P. as
$\begin{array}{l}{a_{n - 1}} = a + \left( {n - 1 - 1} \right)d\\ \Rightarrow {a_{n - 1}} = a + \left( {n - 2} \right)d\end{array}$
Simplifying the expression, we get
$\Rightarrow {a_{n - 1}} = a + nd - 2d$
Substituting $a = 3$ and $d = - 2$, we get
$\Rightarrow {a_{n - 1}} = 3 + n\left( { - 2} \right) - 2\left( { - 2} \right)$
Multiplying the terms, we get

$\Rightarrow {a_{n - 1}} = 3 - 2n + 4$
Adding 3 and 4, we get
$\Rightarrow {a_{n - 1}} = 7 - 2n$
Thus, the ${\left( {n - 1} \right)^{th}}$ term of the A.P. is $7 - 2n$.
We get the value of ${t_{n - 1}}$ as $7 - 2n$. The correct option is option (d).