Let a₁, a₂, a₃, be an arithmetic progression such that its third term is 10 and its sixth term is 19 . Then the value of 4 _ n=1 ^ a_n 3^n is equal to

Step-by-step solution

Let the first term be a₁ and the common difference be d . Given a₃ = a₁ + 2d = 10 and a₆ = a₁ + 5d = 19 . Subtracting the two equations gives 3d = 9 d = 3 . Substituting d = 3 into the first equation gives a₁ + 6 = 10 a₁ = 4 . Let S = _ n=1 ^ a_n 3^n = a₁ 3 + a₂ 3^2 + a₃ 3^3 + This is an infinite arithmetico-geometric progression with common ratio r = 1 3 . The sum of an infinite AGP is given by S = a₁ r 1-r + d r^2 (1-r)^2 . Substituting the values: S = 4 1 3 1 - 1 3 + 3 ( 1 3 )^2 (1 - 1 3 )^2 S = 4/3 2/3 + 3/9 4/