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it's an easy proof....but it's still nice to see some professional do it

i wrote that formula wrong...

See
\[\S_n = \cfrac{n}{2} (2a + (n-1)d)\]

and..?

Now
put
2a +(n-1)d = a + a+(n-1)d = a_1 + a_n

You get ur answer !

i don't think that proves anything...that was formula transformation

see
a_n = a+(n-1)d

That really does //////////// :)

You don't want to use the first formula ?

can you prove a_n = a + (n-1)d?

I think yes

yep

i think you;re confused....this is proving...not substitution

And u asked for a proof ;)

that wasn't a proof @vishweshshrimali5 ....it was formula transformation

and substitution isn't proving either.... @shubhamsrg

Whatever........... leave it
I can prove that without using a_n formula

substitution ? o.O

But you will have to accept that a_n = a+(n-1)d
That is basic def. of AP

I am also surprised @shubhamsrg O.o

i am not! ;)

are any of you familiar with mathematical induction?

You want that proof........

what both of you did was substitution (which is NEVER accepted as proofs)

Let P(n) be the given statement

no. i want other proof

induction is too dull and boring and predictable

you're amazing you know that ?

not you ! ;)

@lgbasallote
please give us a "hint" of the proof u want

actually...it's surprising @vishweshshrimali5 does know induction...

hint? lol..

any proof as long as it's valid and not induction

induction is one of the many great tools we have in mathematics today!

preferrably, i like to see direct proof and contradiction

i like deduction much more

Ok lets try u want to deduce a formula from a continuous pattern

time for the site to go down....

bye

so i suppose i'll just ask this again in the fture

will check out later

yep