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

Geometric Series Quiz 2

mathematics

 If 7, 63, 567 are in a GP, then which two numbers can be inserted between, such that after inserting the number again the obtained number is in series with the GP?

 Calculate the GM between first number and second number and second number and third number.

  $\begin{align} & \text{As the numbers 7, 63, 567 are in GP.}\\ & \text{GM between 7 and 63 } = \sqrt {63 \times 7}  = \sqrt {441}  = 21\\ & \text{GM between 63 and 567 } = \sqrt {63 \times 567}  = \sqrt {35721}  = 189\\ & \text{Hence, two such numbers are 21 and 189.}\end{align}$

 If we insert 4 geometric means between 6 and 1458, then which number will not be the value of inserted GM?

 Calculate the value of all 5 GM which can be inserted between 6 and 1458.

  $\begin{align} & \text{Total number of inserted GM = 4} \\ & \Rightarrow r = {\left( {\dfrac{{1458}}{6}} \right)^{\frac{1}{{4 + 1}}}} \\ &\Rightarrow r = {243^{\frac{1}{5}}} \\ &\Rightarrow {r^5} = {\left[ {{{\left( 3 \right)}^5}} \right]^{\frac{1}{5}}} \\ &\Rightarrow r = 3 \\ & \text{Hence, 324 will not be a value of GM.}\end{align}$

 What will be the value of GM of four terms which are in GP of 40 whose first term is 5 and last term is 320 ?

 First insert 2 more terms in between 5 and 320, then calculate the GM of all four terms.

 $\begin{align} & \Rightarrow r = {\left( {\dfrac{{320}}{5}} \right)^{\frac{1}{{2 + 1}}}} \\ &\Rightarrow r = {64^{\frac{1}{3}}} \\ &\Rightarrow {r^3} = {\left( {{4^3}} \right)^{\frac{1}{3}}} \\ &\Rightarrow r = 4 \\ & \text{Numbers will be 5, 20, 80, 320 }\\ & GM = {\left( {5 \times 20 \times 80 \times 320} \right)^{\frac{1}{4}}} \\ &\Rightarrow GM = 40 \end{align}$ 

 Find the GM value of five terms that are in GP, and the product of all GP terms is 14348907.

 If the total numbers of terms of a GP are odd then the middle term will be GP, so here the value of the third element will be the value of GP.

 Five terms of GP is, $ \dfrac{a}{{{r^2}}} \times \dfrac{a}{r} \times a \times ar \times a{r^2} = 14348907 \\ \Rightarrow {a^5} = 14348907 \\ \Rightarrow a = 27 \\ $So, value of $GM = 27$, as value of GM will be equal to value of middle term i.e. third term.

 If the product of three terms of first GP is 19683. If the first term of second GP is two thirds as the GM of first GP and the product of three terms of second GP is 157464, then what will be the value of GM of second GP?

 From the product of terms of first GP, calculate the GM of first GP, then calculate the first term of second GP, and then with the help of first term of second GP calculate common ratio of second GP.

 $\text{As it is three terms of first GP,  } \dfrac{a}{r} \times a \times ar = 19683 \\ \Rightarrow {a^3} = 19683 \Rightarrow a = 27\\ \text{So, First term of second GP } = \dfrac{2}{3} \times 27 = 18\\ \text{Let common ratio of second GP be “z” }\\ \text{So, }(18 \times 18z \times 18{z^2}) = 157464 \\ \Rightarrow {z^3} = \dfrac{{157464}}{{18 \times 18 \times 18}} \\ \Rightarrow {z^3} = 27 \\ \Rightarrow z = 3 \\  \text{Terms of second }GP = 28, 54, 162\\  \text{Value of GM of second GP = middle term = 54}$