Question

an 11th sequences and series question. don't scam or spam
what if given
$x≠y$
but
$x(a-b) = y(a-b)$
Then?
I concluded this when I was solving question 5 of exercise 9.2
I did
$1=q[a+(n-1)d]=p(a+(n-1)d]$
and got
$p(a-d)=q(a-d)$
but it is already given that
$p≠q$
help me please
​

Answer 1

Answer:

an 11th sequences and series question. don't scam or spam

what if given

$x≠y$

but

$x(a-b) = y(a-b)$

Then?

I concluded this when I was solving question 5 of exercise 9.2

I did

$1=q[a+(n-1)d]=p(a+(n-1)d]$

and got

$p(a-d)=q(a-d)$

but it is already given that

$p≠q$

help me pleasexvbbjjkjkkk

Step-by-step explanation:

an 11th sequences and series question. don't scam or spam

what if given

$x≠y$

but

$x(a-b) = y(a-b)$

Then?

I concluded this when I was solving question 5 of exercise 9.2

I did

$1=q[a+(n-1)d]=p(a+(n-1)d]$

and got

$p(a-d)=q(a-d)$

but it is already given that

$p≠q$

help me pleasevnjhgfghjhgffgggg