Question

find the sum of indicator geometric sequence 1/9, 1/3, 1.....5th term

Answer 1

$the \: series \: \frac{1}{9} , \frac{1}{3} , 1 ... 5th \\ {1}^{st} = \frac{1}{9} \\ {2}^{nd} = 3 \times \frac{1}{9} = \frac{1}{3} \\ {}^{rd} = 3 \times \frac{1}{3} = 1 \\ {4}^{th} = 1 \times 3 = 3 \\ {5}^{th} = 3 \times 3 = 9 \\ and \: so \: on.... \\ so \: answer \: is \: 3.$

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Answer 2

Answer:

Hey mate,

These kind of questions are easy, all you need is to try and understand them

As it is a question from geometric progressions and you need to find the sum of the first n terms of a geometric progression you need to use the sum of definite terms of a geometric progression formula

Therefore,

Sn = [a(r^n -1)]/(r-1)

The answer is in the attachment below please do check it