(B) The $n^{th}$ term of a $G.P.$ is given by $T_n = ar^{n-1}$,where $a$ is the first term and $r$ is the common ratio.Given $T_6 = 32$,we have $ar^5

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(B) The $n^{th}$ term of a $G.P.$ is given by $T_n = ar^{n-1}$,where $a$ is the first term and $r$ is the common ratio.Given $T_6 = 32$,we have $ar^5 = 32$ .....$(i)$Given $T_8 = 128$,we have $ar^7 = 128$ .....$(ii)$Dividing equation $(ii)$ by equation $(i)$:$\frac{ar^7}{ar^5} = \frac{128}{32}$$r^2 = 4$$r = \pm 2$Since the options provided include $2$,the common ratio is $2$.

Competitive Quantitative Aptitude Progression and Sequence Progression and Sequence

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The $6^{th}$ term of a $G.P.$ is $32$ and its $8^{th}$ term is $128$,then the common ratio of the $G.P.$ is

A
$-1$
B
$2$
C
$4$
D
$-4$

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Let $a, b, c, d \in R^+$ such that $256 abcd \geq (a+b+c+d)^4$ and $3a + b + 2c + 5d = 11$. Then,the value of $a^3 + b + c^2 + 5d$ is equal to:

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The sum of $n$ terms of the series $\frac{1}{1 + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{7}} + \dots$ is

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The $3^{rd}$ term of a $G.P.$ is the square of the first term. If the second term is $8,$ determine the $6^{th}$ term.

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If the ${p^{th}}$ term of an $A.P.$ is $q$ and the ${q^{th}}$ term is $p$,then its ${r^{th}}$ term will be

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The $n^{th}$ term of the series $\frac{2}{1!} + \frac{7}{2!} + \frac{15}{3!} + \frac{26}{4!} + \dots$ is

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The $6^{th}$ term of a $G.P.$ is $32$ and its $8^{th}$ term is $128$,then the common ratio of the $G.P.$ is

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Let $a, b, c, d \in R^+$ such that $256 abcd \geq (a+b+c+d)^4$ and $3a + b + 2c + 5d = 11$. Then,the value of $a^3 + b + c^2 + 5d$ is equal to:

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