We appreciate your visit to If the sixth term of an arithmetic progression AP is equal to 3 times the ninth term show that the twelfth term of the AP. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
Let the first term of the arithmetic progression (AP) be [tex]$a$[/tex] and the common difference be [tex]$d$[/tex]. Then, the general term is given by
[tex]$$
T_n = a + (n - 1)d.
$$[/tex]
Thus, the 6th and 9th terms are
[tex]$$
T_6 = a + 5d,\quad T_9 = a + 8d.
$$[/tex]
We are told that 5 times the 6th term is equal to 3 times the 9th term. That is,
[tex]$$
5(a + 5d) = 3(a + 8d).
$$[/tex]
Expanding both sides, we have
[tex]$$
5a + 25d = 3a + 24d.
$$[/tex]
Subtracting [tex]$3a$[/tex] and [tex]$24d$[/tex] from both sides results in
[tex]$$
5a - 3a + 25d - 24d = 0 \quad \Rightarrow \quad 2a + d = 0.
$$[/tex]
This can be rearranged to express [tex]$d$[/tex] in terms of [tex]$a$[/tex]:
[tex]$$
d = -2a.
$$[/tex]
Now, consider the 12th term of the AP:
[tex]$$
T_{12} = a + 11d.
$$[/tex]
Substitute [tex]$d = -2a$[/tex] into the expression for [tex]$T_{12}$[/tex]:
[tex]$$
T_{12} = a + 11(-2a) = a - 22a = -21a.
$$[/tex]
For the 12th term to be equal to the first term, we set
[tex]$$
-21a = a.
$$[/tex]
Solving for [tex]$a$[/tex], subtract [tex]$a$[/tex] from both sides:
[tex]$$
-21a - a = 0 \quad \Rightarrow \quad -22a = 0 \quad \Rightarrow \quad a = 0.
$$[/tex]
Since [tex]$d = -2a$[/tex], it follows that
[tex]$$
d = -2 \cdot 0 = 0.
$$[/tex]
Thus, every term in the AP is
[tex]$$
T_n = a + (n-1)d = 0 + (n-1) \cdot 0 = 0.
$$[/tex]
In particular, the 12th term is
[tex]$$
T_{12} = 0,
$$[/tex]
which is equal to the first term. Hence, we have shown that if [tex]$5T_6 = 3T_9$[/tex], then indeed the 12th term of the AP is equal to the first term.
[tex]$$
T_n = a + (n - 1)d.
$$[/tex]
Thus, the 6th and 9th terms are
[tex]$$
T_6 = a + 5d,\quad T_9 = a + 8d.
$$[/tex]
We are told that 5 times the 6th term is equal to 3 times the 9th term. That is,
[tex]$$
5(a + 5d) = 3(a + 8d).
$$[/tex]
Expanding both sides, we have
[tex]$$
5a + 25d = 3a + 24d.
$$[/tex]
Subtracting [tex]$3a$[/tex] and [tex]$24d$[/tex] from both sides results in
[tex]$$
5a - 3a + 25d - 24d = 0 \quad \Rightarrow \quad 2a + d = 0.
$$[/tex]
This can be rearranged to express [tex]$d$[/tex] in terms of [tex]$a$[/tex]:
[tex]$$
d = -2a.
$$[/tex]
Now, consider the 12th term of the AP:
[tex]$$
T_{12} = a + 11d.
$$[/tex]
Substitute [tex]$d = -2a$[/tex] into the expression for [tex]$T_{12}$[/tex]:
[tex]$$
T_{12} = a + 11(-2a) = a - 22a = -21a.
$$[/tex]
For the 12th term to be equal to the first term, we set
[tex]$$
-21a = a.
$$[/tex]
Solving for [tex]$a$[/tex], subtract [tex]$a$[/tex] from both sides:
[tex]$$
-21a - a = 0 \quad \Rightarrow \quad -22a = 0 \quad \Rightarrow \quad a = 0.
$$[/tex]
Since [tex]$d = -2a$[/tex], it follows that
[tex]$$
d = -2 \cdot 0 = 0.
$$[/tex]
Thus, every term in the AP is
[tex]$$
T_n = a + (n-1)d = 0 + (n-1) \cdot 0 = 0.
$$[/tex]
In particular, the 12th term is
[tex]$$
T_{12} = 0,
$$[/tex]
which is equal to the first term. Hence, we have shown that if [tex]$5T_6 = 3T_9$[/tex], then indeed the 12th term of the AP is equal to the first term.
Thanks for taking the time to read If the sixth term of an arithmetic progression AP is equal to 3 times the ninth term show that the twelfth term of the AP. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!
- Why do Businesses Exist Why does Starbucks Exist What Service does Starbucks Provide Really what is their product.
- The pattern of numbers below is an arithmetic sequence tex 14 24 34 44 54 ldots tex Which statement describes the recursive function used to..
- Morgan felt the need to streamline Edison Electric What changes did Morgan make.
Rewritten by : Barada
