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

If tn=5−2n{t_n} = 5 - 2n, then tn−1={t_{n - 1}} =
(a) 2n−12n - 1
(b) 7+2n7 + 2n
(c) 4−2n4 - 2n
(d) 7−2n7 - 2n

Explanation

Solution

Here, we need to find the value of tn−1{t_{n - 1}}. The equation tn=5−2n{t_n} = 5 - 2n is true for all values of nn. We will substitute the required value of nn 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 nn in the given expression to get the required answer.
It is given that tn=5−2n{t_n} = 5 - 2n.
This is true for all values of nn.
For example, we can find the values of t1{t_1}, t2{t_2}, t3{t_3}, etc.
Substituting n=1n = 1, we get
t1=5−2(1)=5−2=3{t_1} = 5 - 2\left( 1 \right) = 5 - 2 = 3
Similarly, we can find other values like
t2=5−2(2)=5−4=1{t_2} = 5 - 2\left( 2 \right) = 5 - 4 = 1
t3=5−2(3)=5−6=−1{t_3} = 5 - 2\left( 3 \right) = 5 - 6 = - 1
Now, we need to find the value of tn−1{t_{n - 1}}.
We will substitute n−1n - 1 for nn in the expression .
Substituting n−1n - 1 for nn, we get
tn−1=5−2(n−1){t_{n - 1}} = 5 - 2\left( {n - 1} \right)
Multiplying the terms in the expression, we get
⇒tn−1=5−2n+2\Rightarrow {t_{n - 1}} = 5 - 2n + 2
Adding 5 and 2, we get
⇒tn−1=7−2n\Rightarrow {t_{n - 1}} = 7 - 2n
So, we get the value of tn−1{t_{n - 1}} as 7−2n7 - 2n.
The correct option is option (d).

Note: We can also solve this problem using the formula for nth{n^{th}} term of an Arithmetic Progression.
We can find the values of t1{t_1}, t2{t_2}, t3{t_3}, etc.
Substituting n=1n = 1, we get
t1=5−2(1)=5−2=3{t_1} = 5 - 2\left( 1 \right) = 5 - 2 = 3
Similarly, we can find other values like
t2=5−2(2)=5−4=1{t_2} = 5 - 2\left( 2 \right) = 5 - 4 = 1
t3=5−2(3)=5−6=−1{t_3} = 5 - 2\left( 3 \right) = 5 - 6 = - 1
Now, this forms a sequence 3,1,−1,………,(5−2n)3,1, - 1, \ldots \ldots \ldots ,\left( {5 - 2n} \right).
This forms an arithmetic progression with first term a=3a = 3, common difference d=1−3=−2d = 1 - 3 = - 2, and number of terms nn.
We know that nth{n^{th}} term of an A.P. is given by the formula
an=a+(n−1)d{a_n} = a + \left( {n - 1} \right)d, where aa is the first term, dd is the common difference, and nn is the number of terms.
Substituting n−1n - 1 for nn, we get, we get the (n−1)th{\left( {n - 1} \right)^{th}} term of the A.P. as
an−1=a+(n−1−1)d ⇒an−1=a+(n−2)d\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
⇒an−1=a+nd−2d\Rightarrow {a_{n - 1}} = a + nd - 2d
Substituting a=3a = 3 and d=−2d = - 2, we get
⇒an−1=3+n(−2)−2(−2)\Rightarrow {a_{n - 1}} = 3 + n\left( { - 2} \right) - 2\left( { - 2} \right)
Multiplying the terms, we get

⇒an−1=3−2n+4 \Rightarrow {a_{n - 1}} = 3 - 2n + 4
Adding 3 and 4, we get
⇒an−1=7−2n\Rightarrow {a_{n - 1}} = 7 - 2n
Thus, the (n−1)th{\left( {n - 1} \right)^{th}} term of the A.P. is 7−2n7 - 2n.
We get the value of tn−1{t_{n - 1}} as 7−2n7 - 2n. The correct option is option (d).