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This is the second of our many quizzes on geometric progression or GP. Click on this to access more such competitive questions on Geometric Progression.

Geometric Progression Quiz 2

mathematics

 If the GM of three numbers which are in GP is 12, if the two extreme numbers are divided by 2 and their sum is multiplied by the one-fourth of the middle term then we will get 60, what will be the value of common ratio?

 The value of GM will be the value of middle terms, consider $\dfrac{a}{r},a,ar$as terms of GP, then calculate $r$.

  $\text{Three numbers will be, }\dfrac{a}{r},a,ar, \text{ and } GM = 12\\ \text{ So, }\left( {\dfrac{a}{{2r}} + \dfrac{{ar}}{2}} \right) \times \dfrac{a}{4} = 60 \,\,\,\,\,\,\,\text{ As, }(a = 12) \\  \Rightarrow  \left( {\dfrac{{12 + 12{r^2}}}{{2r}}} \right) \times \dfrac{{12}}{4} = 60 \\  \Rightarrow  \left( {\dfrac{{6 + 6{r^2}}}{r}} \right) = 20 \\ \Rightarrow  6{r^2} - 20r + 6 = 0 \\ \Rightarrow  (6r + 2)(r - 3) = 0 \\ \Rightarrow  r =  - 2, + 3$ 

 What will be the product of five terms of a GP if the GM of the progression is 8?

 Value of GM = value of middle terms, consider $\dfrac{a}{{{r^2}}},\dfrac{a}{r},a,ar,a{r^2}$ as terms of GP.

  $\text{Product of all terms of GP will be,}\\  \dfrac{a}{{{r^2}}} \times \dfrac{a}{r} \times a \times ar \times a{r^2} \\ \Rightarrow  {a^5} \\ \Rightarrow  {8^5} \\ \Rightarrow  32768$ 

 Four numbers are in GP, what will be the product of double of first and one-sixth of last terms if the product of all the terms of GP is 5184?

 Consider the terms are $\dfrac{a}{{{r^3}}},\dfrac{a}{r},ar,a{r^3}$, from here convert the required product in terms of $a$.

 $\text{Product of all terms of GP will be, }\\   \dfrac{a}{{{r^3}}} \times \dfrac{a}{r} \times ar \times a{r^3} = 5184 \\ \Rightarrow  {a^4} = 5184 \\ \text{Product of double of first and one-sixth of last terms }= \dfrac{{2a}}{{{r^3}}} \times \dfrac{{a{r^3}}}{6} \\ \Rightarrow  \dfrac{{{a^2}}}{3} = \sqrt {\dfrac{{{a^4}}}{9}}  \\ \Rightarrow  \dfrac{{{a^2}}}{3} = \sqrt {\dfrac{{5184}}{9}}  \\ \Rightarrow  \dfrac{{{a^2}}}{3} = 24$ 

 The sum of three numbers which are in GP is 84, if the first number is doubled and the last number of sequences is halved then the product of these two numbers will become 576, then what will be the value of common ratio?

 Consider $\dfrac{a}{r},a,ar$ as three numbers, from the relationship of first and last term calculate value of $a$.

 $\text{Three numbers will be }\dfrac{a}{r} + a + ar = 84 \\ \Rightarrow  \dfrac{{a + ar + a{r^2}}}{r} = 84 ------\left( 1 \right) \\ \text{Again, } \dfrac{{2a}}{r} \times \dfrac{{ar}}{2} = 576 \\ \Rightarrow  a = \sqrt {576}  = 24 \\ \text{Substitute value of “a” in equation (1),} \\   \dfrac{{24 + 24r + 24{r^2}}}{r} = 84 \\ \Rightarrow  2{r^2} + 2r + 2 = 7r \\ \Rightarrow  2{r^2} - 5r + 2 = 0 \\ \Rightarrow  (2r + 1)(r - 2) = 0 \\ \Rightarrow  r =  - 1,2$ 

 What will be the sum of all three terms of GP if the product of the first two terms is 12 and product of last two terms of GP is 108?

 Consider $\dfrac{a}{r},a,ar$as three terms of GP, with the help of given data convert $r$ in terms of $a$, then calculate the value of $a$.

 $\text{Three numbers will be }\\ \dfrac{a}{r} \times a = 12 \\ \Rightarrow  r = \dfrac{{{a^2}}}{{12}}  ----\left( 1 \right) \\ \text{And, }a \times ar = 108 \\ \Rightarrow  {a^2}\left( {\dfrac{{{a^2}}}{{12}}} \right) = 108  ---\text{from equation }(1)\\ \Rightarrow  {a^4} = 1296 \\ \Rightarrow  {a^4} = {6^4} \\ \Rightarrow  a = 6 \\ \Rightarrow  r = \dfrac{{{a^2}}}{3} = \dfrac{{36}}{{12}} = 3 \\ \text{ Hence, sum of terms }  \dfrac{a}{r} + a + ar = \dfrac{6}{3} + 6 + \left( {6 \times 3} \right) = 26$ 