- The nn𝑛APcap A cap P𝐴𝑃an=a+(n−1)da sub n equals a plus open paren n minus 1 close paren d𝑎𝑛=𝑎+(𝑛−1)𝑑
- The sum of the first nn𝑛APcap A cap P𝐴𝑃Sn=n2(2a+(n−1)d)cap S sub n equals n over 2 end-fraction open paren 2 a plus open paren n minus 1 close paren d close paren𝑆𝑛=𝑛2(2𝑎+(𝑛−1)𝑑)Sn=n2(a+an)cap S sub n equals n over 2 end-fraction open paren a plus a sub n close paren𝑆𝑛=𝑛2(𝑎+𝑎𝑛)
- Here, aa𝑎dd𝑑nn𝑛
- SolutionThe key formulas for sequences and series include those for arithmetic progression (an=a+(n−1)da sub n equals a plus open paren n minus 1 close paren d𝑎𝑛=𝑎+(𝑛−1)𝑑Sn=n2(2a+(n−1)d)cap S sub n equals n over 2 end-fraction open paren 2 a plus open paren n minus 1 close paren d close paren𝑆𝑛=𝑛2(2𝑎+(𝑛−1)𝑑)an=arn−1a sub n equals a r raised to the n minus 1 power𝑎𝑛=𝑎𝑟𝑛−1Sn=a(rn−1)r−1cap S sub n equals the fraction with numerator a open paren r to the n-th power minus 1 close paren and denominator r minus 1 end-fraction𝑆𝑛=𝑎(𝑟𝑛−1)𝑟−1∑k=1nk=n(n+1)2sum from k equals 1 to n of k equals the fraction with numerator n open paren n plus 1 close paren and denominator 2 end-fraction𝑛𝑘=1𝑘=𝑛(𝑛+1)2