IIT JEE Arithmetic Geometric Progression | JEE Arithmetic Geometric

Suppose a₁, a₂, a₃, …. is an A.P. and b₁, b₂, b₃, …… is a G.P. Then the sequence a₁b₁, a₂b₂, …, a_nb_n is said to be an arithmetic-geometric progression. An arithmetic-geometric progression is of the form ab, (a+d)br, (a + 2d)br², (a + 3d)br³, ……

Its sum S_n to n terms is given by

S_n = ab + (a+d)br + (a+2d)br² +……+ (a+(n–2)d)br^n–2 + (a+(n–1)d)br^n–1.

Multiply both sides by r, so that

rS_n = abr+(a+d)br²+…+(a+(n–3)d)br^n–2+(a+(n–2)d)br^n–1+(a+(n–1)d)brⁿ.

Subtracting we get

(1 – r)S_n = ab + dbr + dbr² +…+ dbr^n–2 + dbr^n–1 – (a+(n–1)d)brⁿ.

= ab + dbr(1–r^n–1)/(1–r) (a+(n–1)d)brⁿ

⇒ S_n = ab/1–r + dbr(1–rn–1)/(1–r)2 – (a+(n–1)d)brⁿ/1–r.

If –1 < r < 1, the sum of the infinite number of terms of the progression is

lim_n→∞ S_n = ab/1–r + dbr/(1–r)².

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