In an arithmetic progression,the first term is 1,the n th term is 20 and sum is 399. what is n?

• Ans : 38
  sum = n/2 * (first term + last term )
  so, 399 = n/2 * (20+1)
  hence, n=38
• 11 years agoHelpfull: Yes(43) No(0)
• sum = n/2 * (first term + last term )
  so, 399 = n/2 * (20+1)
  solve both equation's and get n =38.
• 11 years agoHelpfull: Yes(6) No(1)
• given, a=1, l=20 & s=399
  using, s= n/2 * (a+l)
  => 399= n/2 * (1+20)
  => n=38
  
• 11 years agoHelpfull: Yes(3) No(1)
• n(1+nth term 20)/2=399
  so n=399*2/21
  hence n is 38
• 11 years agoHelpfull: Yes(2) No(0)
• Sum of A.P. series -
  s[sum of series]=n[total no. of terms](a1[first term]+an[last term])/2
  putting given values-
  399=n(1+21)/2
  Hence, n=38
• 11 years agoHelpfull: Yes(2) No(0)
• Sum of A.P. series -
  s[sum of series]=n[total no. of terms](a1[first term]+an[last term])/2
  putting given values-
  399=n(1+20)/2
  Hence, n=38
• 11 years agoHelpfull: Yes(1) No(0)
• 38
  399*2/21 by the formula s=(n/2)*(a+l)
• 11 years agoHelpfull: Yes(1) No(1)
• n=38.....,...
• 11 years agoHelpfull: Yes(1) No(0)
• n/2(a+l) by applying that we get n=38
• 11 years agoHelpfull: Yes(1) No(0)
• tn=a+(n-1)d
  a=1,tn=20
  20=1+(n-1)d
  (n-1)d=19
  sn=n/2[2a+(n-1)d]
  399=n/2[2+19]
  n=399*2/21
  n=38
• 11 years agoHelpfull: Yes(1) No(0)
• first term=1, last term=20, sum=399
  sum=n/2(first+last terms)
  399=n/2(1+20)
  n=(399/21)*2
  n=38
• 10 years agoHelpfull: Yes(0) No(0)