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.
