Find the sum of the G.P.:

11/5, 11/25, 11/125, 11/625 , ... to n terms.

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  1. 2/5(1-(1/5)n)
  2. 11/4(1-(1/5)n)
  3. 11/5(1-(1/5)n)
  4. 4/11(1-(1/5)n)

Answer (Detailed Solution Below)

Option 2 : 11/4(1-(1/5)n)
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यूपी पुलिस SI (दरोगा) सामान्य हिंदी मॉक टेस्ट
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Detailed Solution

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Given:

11/5, 11/25, 11/125, 11/625 , ... to n terms.

Formula used:

Sum = a (1 - rn) / (1 - r )

Where a = first term & r = common difference 

Calculation:

Common difference (r) = (11/25) ÷ (11/5)

⇒ (1/5)

Putting the value of r in the above formula we get:

Sum = \(\frac{11}{5}\)(1 - \(\frac{1}{5}\)n) / 1 - \(\frac{1}{5}\)

⇒ \(\frac{11}{5}\)(1- (\(\frac{1}{5}\))n) / \(\frac{4}{5}\)

⇒ \(\frac{11}{4}\)(1 - (\(\frac{1}{5}\))n)

Hence the correct answer is "​​​\(\frac{11}{4}\)(1 - (​\(\frac{1}{5}\))n)".

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