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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.

**Complete step-by-step 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).

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