The sum of third and ninth term of an A.P is 8. Find the sum of the first 11 terms of the progression.

44
22
19
None of the above

• The third term t3 = a + 3d
  The ninth term t9 = a + 8d
  t3 + t9 = 2a + 10d = 8
  
  Sum of first 11 terms of an AP is given by
  =11/2(2a+10d)
  =11/2(8)
  =88/2
  =44
• 10 years agoHelpfull: Yes(29) No(5)
• a3 + a9 = 8
  [a1 + (3-1)d] + [a1+ (9-1)d] = 8
  2a1 + 10d = 8------------------(1)
  
  Sum of first 11 terms-
  S11= 11/2 * [2a1+(11-1)d]
  S11= 11/2 * 8-----------(from 1)
  S11=44
• 10 years agoHelpfull: Yes(14) No(1)
• Third term = a + 2d
  ninth term = a + 8d
  sum of two terms is equal to 8 i.e..., 2a+10d = 8
  take two common i.e.., a + 5d =4 ------- (1)
  first 11 terms is a,a+d,a+2d,a+3d,a+4d,a+5d,a+6d,a+7d,a+8d,a+9d,a+10d.
  sum of the first 11 terms is 11a+55d
  take 11 common 11(a+5d)----->(2)
  sub (1) in (2)
  we get answer 44
• 8 years agoHelpfull: Yes(3) No(1)
• Formula
  Sum=n/2 (2a + (n-1)d)
  therefore , 3/2 (2a + (3-1)d) = 9/2 (2a+(9-1)d)
  =3/2 (2a + 2d) = 9/2 (2a + 8d)
  =3a+3d = 9a + 36d
  => 6a = - 33d
  => a = - 11/2 d
  
  Now Sum of 11th term = 11/2 ( 2a + (n-1) d)
  = 11/2 (2 x (-11/2)d + 10 d)
  = 11/2 (-11d + 10d )
  = 11/2 (-d) (Ans)
• 10 years agoHelpfull: Yes(0) No(6)
• Sum=n2(a+l) as a+l=8(given)
  For 3 to 9 sum is (7/2(8)=28)
  22 19 two r lesser than 28 .
  so 44 answer.
• 8 years agoHelpfull: Yes(0) No(0)