how many terms of the AP 9,7,25..... must be taken to give a sum of 63

1.	how many terms of the AP 9,7,25..... must be taken to give a sum of 636
Answer» it ll be 17 in place of 7.Ans. First Term (a) = 9Common Difference (d) = 17-9 = 8\xa0Sn\xa0= 636We Know\\(=> S_n = {n\\over 2}[2a+(n-1)d]\\)\\(=> 636= {n\\over 2}[2\\times 9 +(n-1)8]\\)\\(=> 636= {n\\over 2}[18 +8n-8]\\)\\(=> 636= {n\\over 2}[10 +8n] => 636=n[5 +4n]\\)=> 4n2 + 5n - 636 = 0=> 4n2 + 53n - 48n - 636 = 0=> n (4n+53) - 12(4n+ 53) = 0=> (4n+53) (n-12) = 0Either 4n + 53 = 0 or n -12 = 0n = (-53\\4) or 12\xa0Number of terms = 12\xa0

Discussion