Answer :

Answer:

Step-by-step explanation:

formulae for AP

sum of n terms in AP, Sn = [tex]\frac{n}{2\\}[/tex] [2a+(n-1)d]

nth term of AP, aₙ=a₁ + (n-1) d

let there be 6 terms in an AP

the first term be a₁ and so the difference be d

then the 5th term from the beginning will be a₅=a₁ + (5-1) d = a₁ + 4d

and 5th term from the last i.e a₂=a₁+d

sum of 5th term from beginning and 5th term from the last be (a₅ and a₂ respectively), = (a₁ + 4d)+(a₁+d) = 2a₁+5d

sum of 1st and last term is (a₁+a₆) =a₁+(a₁ + 5d) = 2a₁+5d

hence proved.