Find the sum of the G.P. :

3/5 , 3/25 , 3/125 , 3/625 ,... to n terms.

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UP Police SI (दरोगा) Official PYP (Held On: 29 Nov 2021 Shift 3)
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  1. 3/4(1-(1/5n))
  2. 4/5(1-(1/5n))
  3. 5/4(1-(1/5n))
  4. 2/5(1-(1/5n))

Answer (Detailed Solution Below)

Option 1 : 3/4(1-(1/5n))
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UP Police SI (दरोगा) Official PYP (Held On: 2 Dec 2021 Shift 1)
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Detailed Solution

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Given:

The G.P. series is: 3/5, 3/25, 3/125, 3/625, ... to n terms.

First term (a) = 3/5

Common ratio (r) = 1/5

Number of terms (n) = n

Formula used:

Sum of G.P. for n terms, Sn = a \(\frac{1-r^n}{1-r}\), where |r| < 1

Calculation:

Sn = (3/5) \(\frac{1-(1/5)^n}{1-(1/5)}\)

⇒ Sn = (3/5) \(\frac{1-(1/5)^n}{4/5}\)

⇒ Sn = (3/5) × (5/4) × (1 - (1/5)n)

⇒ Sn = 3/4 (1 - (1/5)n)

∴ The correct answer is option 1.

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