Question

help me to solve these 2 AP's (Arithmetic progression's) questions.​

Answer 1

1. if a,b,c,d and e form an AP then

b=a+D        (D=common difference)

c=a+2D

d=a+3D

e=a+4D

now the given expression is:- a-4b+6c-4d+e

substituting the value of b,c,d and e from above equations

we get,

      a-4(a+D)+6(a+2D)-4(a+3D)+a+4D

    =8a-8a

    =0 ANSWER

2. The arithmetic mean between a and b is =(a+b)/2

given am between a and b= [a^(n+1) + b^(n+1)]/a^n+b^n

  so now,  [a^(n+1) + b^(n+1)]/a^n+b^n =a+b/2

                 

     2[(a^(n+1) + b^(n+1)]=(a+b)(a^n+b^n)

     2a^(n+1) +2b^(n+1)=a.a^n+a.b^n+b.a^n+b.b^n

    by solving this equation we get

      (a-b)(a^n-b^n)=0

     a^n-b^n=0

   a^n=b^n

   (a/b)^n=1    (as we know that x^0=1)

 s0,   (a/b)^n=(a/b)^0

        ∴n=0 ANSWER