If a₁, a₂, a₃ . are in A.P. and b_k=a_k+a_ k+1 + .+a_ k+n-1 (k=1,2,3 . ) , then b₁+b₂+ . .+b_n is equal to

Step-by-step solution

aligned & We have b_k= n 2 (a_k+a_ n+k-1 )= n 2 [a₁+(k-1) d+a₁+(n+k-2) d ] & = n 2 [a_n+a₁+2(k-1) d ] (d is the common difference) & _ k=1 ^n b_k= n 2 [n a_n+n a₁+2 d _ k=1 ^n(k-1) ]= n 2 [n a_n+n a₁+d(n-1) n ] & = n^2 2 [a_n+a₁+(n-1) d ]= n^2 2 [a_n+a_n ]=n^2 a_n aligned