Question 1 of 14beginner🔧 ApplyNumerical2 marks

Find the sum to infinity in each of the following Geometric Progression. 1,13,19,…1, \frac{1}{3}, \frac{1}{9}, \ldots

Correct Answer

1.5

Exercise: Exercise 8.3 | Q: 1 | (Page Page 2)
For More Understanding

Explanation

The question asks for the sum to infinity of a Geometric Progression. According to the provided text, for a G.P. with first term 'a' and common ratio 'r', the sum to infinity exists if |r| < 1 and is given by the formula S = a / (1 - r). The text confirms this limit behavior by explaining that as n approaches infinity, rnr^{n} approaches 0, leaving the sum as a / (1 - r).

Solution Steps

  1. Step 1: Identify the first term and common ratio. The given G.P. is 1,13,19,…1, \frac{1}{3}, \frac{1}{9}, \ldots Here, the first term (a) = 1. The common ratio (r) is found by dividing the second term by the first term: r=1/31=13r = \frac{1/3}{1} = \frac{1}{3}.

  2. Step 2: Check the condition for sum to infinity. Since ∣r∣=∣13∣<1|r| = \left|\frac{1}{3}\right| < 1, the sum to infinity exists.

  3. Step 3: Apply the formula for sum to infinity. The formula is S=a1−rS = \frac{a}{1-r}. Substituting the values of aa and rr:

    S=11−13S = \frac{1}{1 - \frac{1}{3}}

  4. Step 4: Simplify the expression.

    S=133−13=123S = \frac{1}{\frac{3}{3} - \frac{1}{3}} = \frac{1}{\frac{2}{3}} S=1×32=32=1.5S = 1 \times \frac{3}{2} = \frac{3}{2} = 1.5
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