Question

In arithmetic sequence is given:

\[\LARGE a_p=q \]
\[\LARGE a_q=p \]

Find
\[\LARGE a_n=? \]

\[\LARGE A)\quad p-q+n\]
\[\LARGE B)\quad p-q-n\]
\[\LARGE C)\quad-p+q+n\]
\[\LARGE C)\quad p+q-n\]

I can't even start it, leave alone to do it !!!

Answer 1

Hmm.
\[ a_{p} = a _{1}+(p-1)d=q\]
\[ a_{q} = a _{1}+(q-1)d=p\]
\[ a_{p} -a_{q}=(p-q)d=q-p\]
from here d=-1
so:\[a _{n}=a _{1}-n+1\]
But i can't find the conection with the options you have....

Answer 2

@Kreshnik

Answer 3

@myko Thank you very much. You Gave me a great help. !

\[\LARGE a_p=q -------[1]\]
\[\LARGE a_q=p -------[2]\]
\[\LARGE a_p=a_1+(p-1)d---[3] \]
\[\LARGE a_q=a_1+(q-1)d----[4] \]
Now from [1] and [3] if:
\[\LARGE a_p=q \quad \quad and \quad \quad a_p=a_1+(p-1)d \quad \Longrightarrow \]
\[\LARGE q=a_1+(p-1)d-----[5] \]
And from [2] and [4] if:
\[\LARGE a_q=p \quad \quad \quad and\quad \quad a_q=a_1+(q-1)d \quad \Longrightarrow\]

\[\LARGE p=a_1+(q-1)d -----[6]\]

You did well, from
\[\LARGE a_p-a_q=[a_1+(p-1)d]-[a_1+(q-1)d]]= \]
\[\LARGE =(p-1)d-(q-1)d=d[(p-1)-(q-1)]= \]
\[\LARGE d(p-q) \Longrightarrow a_p-a_q=d(p-q) .\]
And from [1] & [2] we know that:
\[\LARGE a_p-a_q=q-p . \]
\[ \LARGE \text{So:}\;\;\; d(p-q)=q-p \quad \Longleftrightarrow d=-1 \]
Now we substitute d=-1 to the main formula: (This is what you've done, I just rewrote them longer :D)
But before we substitute... from:
[1] and [3]
I did.

\[\LARGE q=a_1+(p-1)d \quad \Longrightarrow a_1=q-(p-1)d \]
since d=-1 ...
\[\LARGE a_1=q-(p-1)d=q-(-p+1) \quad \Longrightarrow \]
\[\LARGE a_1=q+p-1 \]
And also from:
[2] & [4]



\[\LARGE p=a_1+(q-1)d \Longrightarrow a_1=p-(q-1)d \]
\[ \LARGE \Longrightarrow a_1=p+q-1 \]

so we get the same result...
and now let's substitute

\[\LARGE a_n=a_1+(n-1)d \Longrightarrow a_n=(p+q-1)+(1-n) \]
\[\LARGE a_n=p+q-n \]

I think this should be correct O_O ...

Answer 4

can someone verify it .. if this is correct or not :O

Answer 5

good job

Answer 6

you too ! :)

Answer 7

an = p+q-n is correct

Answer 8

thank you .