Question

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Answer 1

Explanation:

Answer:

let a be the first term of A.P. & d is the common difference . Then, b = a+d , c= a+2d, l= a+3d, m=a+4d

Now , a-4b+6c-4l+m

putting the value of b,c,l,m

=a -4(a+d)+6(a+2d)-4(a+3d)+a+4d

=a-4a-4d+6a+12d-4a-12d+a+4d

=0.

Answer 2

Let the common difference and number of terms of A.P. be d and n respectively.

∴Last term of A.P.=nth of the A.P.,

anan=4l (Given)2a+(n−1)d=4ln=4l−2ad+1

  ...(1)

Sum of n terms of A.P=S   (Given)

∴n2[4a+(n−1)d]

=S12(4l−2ad+1 )[4a+(4l−2ad+1 −1)d]

=S(2l−ad+1)[4a+(4l−2ad)d]=S(2l−ad+1)[4a+4l−2a]=S2(2l−ad+1)(a+2l)=

S(2l−ad+1

)=S2(a+2l)2l−ad

=S2(a+2l)−12l−ad

=S−2a−4l2a+4ld

=2(2l−a)(2l+a)S−2a−4ld

=2(4l2−a2)S−2a−4l