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, …, anbn 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 Sn to n terms is given by

Sn = ab + (a+d)br + (a+2d)br2 +……+ (a+(n–2)d)brn–2 + (a+(n–1)d)brn–1.

Multiply both sides by r, so that

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

Subtracting we get

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

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

⇒ Sn = 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

limn→∞ Sn = ab/1–r + dbr/(1–r)2.

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