If in a geometric progression left{ {{aₙ}} right},{a₁} = 3,{aₙ} = 96 and {Sₙ} = 189 then val

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8. Sequences and Series

easy

If in a geometric progression $\left\{ {{a_n}} \right\},\;{a_1} = 3,\;{a_n} = 96$ and ${S_n} = 189$ then the value of $n$ is

A

$5$

B

$6$

C

$7$

D

$8$

Solution

(b) ${a_1} = 3,\;{a_n} = 96$

$ \Rightarrow $${a_1}{r^{n – 1}} = 96$

$ \Rightarrow $${r^{n – 1}} = 32$ …..(i)

${S_n} = \frac{{{a_1}({r^n} – 1)}}{{r – 1}} = 189$

$ \Rightarrow $$\frac{{3(32r – 1)}}{{r – 1}} = 189$

Hence $r = 2$ and $n = 6$.

Standard 11

Mathematics

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