Question

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

Answer 1

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).