In an A.P.,s₃=6,s₆=3,prove that 2(2n+1)S_{n +4}=(n+4)S_2n+1.

S₃= 3/ 2(2a + 2d) = 3a + 3d = 6 --->(1)

 S₆ = 6/2 (2a + 5d )= 6a + 15d = 3  ---->(2)

  solving (1) & (2) ....

 a = 3 & d=-1

 now 2(2n+1)s_n+4= n+4 S_2n+1)

2(2n+1)  n+4 /2(2a + (n+4 -1)d) = (n+4) (2n+ 1)/2 (2a + (2n+1-1)d

(2n +1) (n+4) (6 + (n+3)-1) = (n+4)(2n+1)/2 ( 6 - 2n )

2n+1)(n+4)(3-n)  = (n+4)(2n +1)(3-n)  PROVED