Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# If S Be the Sum, P the Product and R Be the Sum of the Reciprocals of N Terms of a Gp, Then P2 is Equal to - Mathematics

MCQ

If S be the sum, P the product and R be the sum of the reciprocals of n terms of a GP, then P2 is equal to

• (a) S/R

• (b) R/S

• (c) (R/S)n

• (d) (S/R)n

#### Solution

(d) $\left( \frac{S}{R} \right)^n$

$\text{ Sum of n terms of the G . P } . , S = \frac{a\left( r^n - 1 \right)}{\left( r - 1 \right)}$
$\text{ Product of n terms of the G . P } . , P = a^n r^\left[ \frac{n\left( n - 1 \right)}{2} \right]$
$\text{ Sum of the reciprocals of n terms of the G . P } . , R = \frac{\left[ \frac{1}{r^n} - 1 \right]}{a\left( \frac{1}{r} - 1 \right)} = \frac{\left( r^n - 1 \right)}{a r^\left( n - 1 \right) \left( r - 1 \right)}$
$\therefore P^2 = \left\{ a^2 r^\frac{2\left( n - 1 \right)}{2} \right\}^n$
$\Rightarrow P^2 = \left\{ \frac{\frac{a\left( r^n - 1 \right)}{\left( r - 1 \right)}}{\frac{\left( r^n - 1 \right)}{a r^\left( n - 1 \right) \left( r - 1 \right)}} \right\}^n$
$\Rightarrow P^2 = \left\{ \frac{S}{R} \right\}^n$
$\text{ Let the first term of the G . P . be a and the common ratio be r } .$
$\text{ Sum of n terms }, S = \frac{a\left( r^n - 1 \right)}{r - 1}$
$\text{ Product of the G . P } . , P = a^n r^\frac{n\left( n + 1 \right)}{2}$
$\text{ Sum of the reciprocals of n terms }, R = \frac{\left( \frac{1}{r^n - 1} \right)}{a\left( \frac{1}{r^{} - 1} \right)} = \frac{\left( \frac{1 - r^n}{r^n} \right)}{a\left( \frac{1 - r}{r} \right)}$
$p^2 = \left\{ a^2 r^\frac{\left( n + 1 \right)}{2} \right\}^n$
$p^2 = \left\{ \frac{\frac{a\left( r^n - 1 \right)}{r - 1}}{\frac{\left( \frac{1 - r^n}{r^n} \right)}{a\left( \frac{1 - r}{r} \right)}} \right\}^n = \left\{ \frac{S}{R} \right\}^n$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 20 Geometric Progression
Q 6 | Page 57