the sum of the terms of an Infinity decreasing geometric progression is 3 and the sum of the cubes of is terms is 108/13 write the progression
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the sum of the terms of an Infinity decreasing geometric progression is 3 and the sum of the cubes of is terms is 108/13 write the progression
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Let the GP be a,ar,ar2,... where 0<r<1.
Then a+ar+ar2+...=3 and a2+a2r2+a2r4+...=29
⇒1−ra=3 and 1−r2a2=29
We have 1−ra=3⇒a=3(1−r)
∴1−r29(1−r)2=29
⇒1+r1−r=21
⇒r=31
Putting r=