Medium MCQ +4 / -1 PYQ · JEE Mains 2022

Let $$S = 2 + {6 \over 7} + {{12} \over {{7^2}}} + {{20} \over {{7^3}}} + {{30} \over {{7^4}}} + \,.....$$. Then 4S is equal to

  1. A ${\left( {{7 \over 3}} \right)^2}$
  2. B ${{{7^3}} \over {{3^2}}}$
  3. C ${\left( {{7 \over 3}} \right)^3}$ Correct answer
  4. D ${{{7^2}} \over {{3^3}}}$

Solution

<p>$$S = 2 + {6 \over 7} + {{12} \over {{7^2}}} + {{20} \over {{7^3}}} + {{30} \over {{7^4}}} + $$ ..... ...... (i)</p> <p>$${1 \over 7}S = {2 \over 7} + {6 \over {{7^2}}} + {{12} \over {{7^3}}} + {{20} \over {{7^4}}} + $$ .... ....... (ii)</p> <p>(i) - (ii)</p> <p>${6 \over 7}S = 2 + {4 \over 7} + {6 \over {{7^2}}} + {8 \over {{7^3}}} +$ ...... ....... (iii)</p> <p>${6 \over {{7^2}}}S = {2 \over 7} + {4 \over {{7^2}}} + {6 \over {{7^3}}} +$ ..... ......... (iv)</p> <p>(iii) - (iv)</p> <p>$${\left( {{6 \over 7}} \right)^2}S = 2 + {2 \over 7} + {2 \over {{7^2}}} + {2 \over {{7^3}}} + $$ ......</p> <p>$= 2\left[ {{1 \over {1 - {1 \over 7}}}} \right] = 2\left( {{7 \over 6}} \right)$</p> <p>$\therefore$ $4S = 8{\left( {{7 \over 6}} \right)^3} = {\left( {{7 \over 3}} \right)^3}$</p>

About this question

Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression

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