Q. The sum of the squares of the fifth and the eleventh term of an AP is 3 and the product of the second and fourteenth term is equal to P.Find the product of the first and the fifteenth term of the AP.

Option
a) (58P-39)/45
b) (98P+39)/72
c) (116P-39)/90
d) (98P+39)/90
e) None of these

• option c
  
  (T5)^2+(T11)^2=3
  
  T2*T14=P
  
  T1*T15=?
  
  (a+4d)^2+(a+10d)^2=3
  2a^2+28ad+116d^2=3.....(1)
  
  (a+d)(a+13d)=P
  
  a^2+14ad+13d^2=P
  
  a^2+14ad = P- 13d^2
  
  1.5-45d^2=P
  
  2P=3-90d^2...........(2)
  
  a^2+14ad=?
  
  =P-13d^2
  
  =P-13(3-2P)/90
  
  =(116P-39)/90
  
  option (C)
• 12 years agoHelpfull: Yes(9) No(4)
• Good job, Dipin
• 12 years agoHelpfull: Yes(1) No(1)