the sum of 3rd and 9th term of A.P is 8. find the sum of 1st 11 terms of A.P
a)33 b)22 c)44 d)55

• A.P. series is a, a+d, a+2d, a+3d, a+4d,..............nth terms
  
  where a = first term
  d = common difference
  
  We have to given,
  sum of 3rd and 9th terms = 8
  (a+2d)+(a+8d) = 8
  2a+10d = 8
  a+5d = 4.............(1)
  
  formula, Sum of 1st 11 terms of AP
  S = n[ 2a+(n-1)d ]/2
  =11[ 2a+(11-1)d ]/2
  =11[a+5d]
  =11[4] ( from eq. ...(1) )
  S = 44 ans.
• 9 years agoHelpfull: Yes(17) No(0)
• (a+2d)+(a+8d)=8
  2a+10d=8,
  now, sum=11/2[2a+(11-1)d]
  =11/2*8
  
  c) 44
• 9 years agoHelpfull: Yes(2) No(0)
• ans.b
  Explanation:S11=11/n(2a+10d)
  
• 9 years agoHelpfull: Yes(0) No(0)
• 55 option d
• 9 years agoHelpfull: Yes(0) No(1)
• c) 44 is d correct ans
• 9 years agoHelpfull: Yes(0) No(0)
• 44 is the correct answer
• 9 years agoHelpfull: Yes(0) No(0)
• 3rd term = a+2d
  9th term = a+8d
  sum =2a+10d
  2a+10d=8
  sum of 11 terms= (n(2a+(n-1)d))/2
  so sum = (11*8)/2
  answer =44
• 9 years agoHelpfull: Yes(0) No(0)
• A/Q t3+t9=8
  (a+2d)+ (a+8d)=8
  2a+10d=8
  Sn=n*(2a+(n-1)d)/2
  S11=11(2a+10d)/2
  =11*8/2
  =44 Ans
  
• 9 years agoHelpfull: Yes(0) No(0)
• Ans: 44
  
  t3=a+2d
  t9=a+8d
  
  t3+t9=2a+10d
  
  sum of 11 terms= 11/2*(2a+10d)
  s11= 11/2*8 =44
• 9 years agoHelpfull: Yes(0) No(0)
• 44
  as terms r like -1,0,1,2,3,4,5,6,7,8,9
  and by summing 11terms we get 44
• 9 years agoHelpfull: Yes(0) No(0)
• t3 + t9 =8 ; (a+2d)+(a+8d)=8; 2a+10d=8; s11=11/2*(2a+10d)=11/2*8=44
• 9 years agoHelpfull: Yes(0) No(0)
• 3rd term is a+2d
  9th term is a+8d
  sum of 3rd and 9th term is 8
  i.e 2a+10d=8
  sum first 11 terms is=n/2*(2a+(n-1)*d)
  =11/2*(2a+10d)
  =11/2*8
  =44
• 9 years agoHelpfull: Yes(0) No(0)