Question

If 64, 27, 36 are the ${{P}^{th}},{{Q}^{th}},{{R}^{th}}$ terms of a GP, then P + 2Q is equal to
(a) R
(b) 2R
(c) 3R
(d) 4R

Answer 1

Hint: First of all, write the ${{P}^{th}},{{Q}^{th}},{{R}^{th}}$ terms by using the general term of G.P that is $a{{r}^{n-1}}$. Now, equate it with the given values and the square the ${{Q}^{th}}$ term and multiply the expression by the expression for the ${{P}^{th}}$ term. Now, write the constant terms of the expression in terms of R to get the values of P + 2Q in terms of R.

Complete step by step solution:
We are given that 64, 27, 36 are the ${{P}^{th}},{{Q}^{th}},{{R}^{th}}$ terms of a GP respectively. We have to find the value of P + 2Q in terms of R. We are given that in a GP or geometrical progression,

{{\text{P}}^{\text{th}}}\text{ term}=64\ldots ..\left( i \right) \\\ {{\text{Q}}^{\text{th}}}\text{ term}=27\ldots ..\left( ii \right) \\\ {{\text{R}}^{\text{th}}}\text{ term}=36\ldots ..\left( iii \right) \\\ \end{array}$$ We know that $${{n}^{th}}$$ term of GP or geometric progression is given by $$a{{r}^{n-1}}$$ where a and r are the first term and common ratio of GP respectively. By substituting the value of n = P, we get, $${{\text{P}}^{\text{th}}}\text{ term}=a{{r}^{P-1}}....\left( iv \right)$$ By substituting the value of n = Q, we get, $${{\text{Q}}^{\text{th}}}\text{ term}=a{{r}^{Q-1}}....\left( v \right)$$ By substituting the value of n = R, we get, $${{\text{R}}^{\text{th}}}\text{ term}=a{{r}^{R-1}}....\left( vi \right)$$ By equating $${{P}^{th}}$$ term from equation (i) and (iv), we get, $$a{{r}^{P-1}}=64....\left( vii \right)$$ By equating $${{Q}^{th}}$$ term from equation (ii) and (v) and squaring both the sides, we get, $${{\left( a{{r}^{Q-1}} \right)}^{2}}={{\left( 27 \right)}^{2}}....\left( viii \right)$$ By equating $${{R}^{th}}$$ term from equation (iii) and (vi), we get, $$a{{r}^{R-1}}=36....\left( ix \right)$$ By multiplying equation (vii) and (viii), we get, $$\left( a{{r}^{P-1}} \right){{\left( a{{r}^{Q-1}} \right)}^{2}}=\left( 64 \right){{\left( 27 \right)}^{2}}$$ We know that $${{\left( ab \right)}^{x}}={{a}^{x}}.{{b}^{x}}$$ So, we get, $$\left( a{{r}^{P-1}} \right).{{a}^{2}}{{\left( {{r}^{Q-1}} \right)}^{2}}=\left( 64 \right){{\left( 27 \right)}^{2}}$$ We know that, $${{a}^{x}}.{{a}^{y}}={{a}^{x+y}}\text{ and }{{\left( {{\left( a \right)}^{x}} \right)}^{y}}={{a}^{xy}}$$ By using this, we get, $$\left( {{a}^{1+2}} \right){{r}^{\left( P-1 \right)+2\left( Q-1 \right)}}=\left( 64 \right){{\left( 27 \right)}^{2}}$$ $$\left( {{a}^{3}} \right)\left( {{r}^{P+2Q-2-1}} \right)=46656$$ We know that $${{\left( 36 \right)}^{3}}=46656$$ By using this, we can write the above equation as, $${{\left( a \right)}^{3}}\left( {{r}^{P+2Q-3}} \right)={{\left( 36 \right)}^{3}}$$ From equation (ix), by writing 36 in terms of a, r and E in the above equation, we get, $$\left( {{a}^{3}} \right)\left( {{r}^{P+2Q-3}} \right)={{\left( a{{r}^{R-1}} \right)}^{3}}$$ $$\left( {{a}^{3}} \right)\left( {{r}^{P+2Q-3}} \right)={{a}^{3}}{{r}^{3\left( R-1 \right)}}$$ $${{r}^{P+2Q-3}}={{r}^{3R-3}}$$ By comparing the powers of r, we get, $$P+2Q-3=3R-3$$ $$P+2Q=3R$$ So, we get the value of P + 2Q as 3R. Hence, the option (c) is the right answer. Note: In this question, many students make this mistake of writing $${{P}^{th}}$$ term as $$a{{r}^{P}},{{Q}^{th}}$$ term as $$a{{r}^{Q}}$$ and so on which is wrong because our general term is $$a{{r}^{n-1}}\text{ not }a{{r}^{n}}$$. So this must be taken care of. Also, in the above question, we square the $${{Q}^{th}}$$ term already because we had to find the value of P + 2Q to get 2Q, we squared the $${{Q}^{th}}$$ term. So, students must know this as well.