Question

Hi guyss
I wanna know everything about the nth term in Arithemetic progression
if u know then answer it clearly​

Answer 1

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Answer 2

$\red{\underline{\underline{Answer:}}}$

$\bold\blue{Explanation:}$

$\sf{n^{th} \ term \ can \ be \ find \ by \ substituting}$

$\sf{values \ of \ first \ term(a), \ Common \ difference(d)}$

$\sf{and \ value \ of \ n \ in \ formula}$

$\sf{tn=a+(n-1)d}$

$\sf{Also, \ in \ some \ sums \ value \ of \ n \ is \ found}$

$\sf{By \ substituting \ given \ values \ in \ formula}$

$\sf{Sn=\frac{n}{2}[2a+(n-1)d]}$

$\sf{Or}$

$\sf{Sn=\frac{n}{2}[t1+tn]}$

$\sf\green{Extra \ information}$

$\boxed{\sf{2\times \ t2=t1+t3}}$

$\sf{n \ is \ always \ natural \ number. }$

$\sf{Conversion}$

$\sf{Sn=\frac{n}{2}[2a+(n-1)d]}$

$\sf{Here, \ 2a+(n-1)d \ can \ also \ written \ as}$

$\sf{a+a+(n-1)d}$

$\sf{we, \ know \ t1=a \ and \ tn=a+(n-1)d}$

$\sf{\therefore{Sn=\frac{n}{2}[t1+tn]}}$