11th

5 10 20 40 80 160 320 640 1280 2560 5120

From geometric progression,

Gn =G_1 r^(n-1).............................1

but Gn = 5120 G1 = 5 r = 10÷5 =2

5120 = 5 * 2^(n-1)

2^(n-1) = 5120\5 = 1024 = 2^(10)

i.e. n - 1 = 10

so n = 11

so it is 11th term.

Given the geometric progression: a 1 = 5 a 2 = 5 x 2 = 10 a 3 = 10 x 2 = 20 or a 1 x 2^2 = 5 x 4 = 20 Then, we can write: a n = a1 x 2 ^(n - 1) being n the term of the progression Then, an = 5120 = 5 x 2 ^(n-1) Solving this equation For n= 11 , 2 ^10 = 1024 , 1024 x 5 = 5120 The term is n= 11

Clearly: The common ration = 10 / 5 = 2. therefore, a = 5; r = 2; As, we know,

a*r^n-1=5120 :5*2^n-1 = 5120 :2^n-1 = 1024 :2^n-1 = 2^ 10 : n - 1= 10 n = 11

First and last term of a geometric progression are 3 and 96. If the sum of all these terms is 189, then find the number of terms in this progression.

The sum of the first three terms of a geometric progression is 8. The sum of the first six terms of the same geometric progression is 12.

Find the common ratio of this geometric progression?