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If the fifth term of a GP is 81 and first term is 16, what will be the 4^th term of the GP?

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} 5^{t h} term of GP \\ = a r^{5 - 1} \\ = 16 \times r^{4} \\ = 81 \\ Or, r = {(\frac{81}{16})}^{\frac{1}{4}} = \frac{3}{2} \\ 4^{t h} term of GP \\ = a r^{4 - 1} \\ = 16 \times {(\frac{3}{2})}^{3} \\ = 54 \end{aligned}

$\eqalign{ & {5^{th}}\,{\text{term}}\,{\text{of}}\,{\text{GP}} \cr & = a{r^{5 - 1}} \cr & = 16 \times {r^4} \cr & = 81 \cr & {\text{Or}},\,r = {\left( {\frac{{81}}{{16}}} \right)^{\frac{1}{4}}} = \frac{3}{2} \cr & {4^{th}}\,{\text{term}}\,{\text{of}}\,{\text{GP}} \cr & = a{r^{4 - 1}} \cr & = 16 \times {\left( {\frac{3}{2}} \right)^3} \cr & = 54 \cr}$

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", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} 5^{t h} term of GP \\ = a r^{5 - 1} \\ = 16 \times r^{4} \\ = 81 \\ Or, r = {(\frac{81}{16})}^{\frac{1}{4}} = \frac{3}{2} \\ 4^{t h} term of GP \\ = a r^{4 - 1} \\ = 16 \times {(\frac{3}{2})}^{3} \\ = 54 \end{aligned}

$\eqalign{ & {5^{th}}\,{\text{term}}\,{\text{of}}\,{\text{GP}} \cr & = a{r^{5 - 1}} \cr & = 16 \times {r^4} \cr & = 81 \cr & {\text{Or}},\,r = {\left( {\frac{{81}}{{16}}} \right)^{\frac{1}{4}}} = \frac{3}{2} \cr & {4^{th}}\,{\text{term}}\,{\text{of}}\,{\text{GP}} \cr & = a{r^{4 - 1}} \cr & = 16 \times {\left( {\frac{3}{2}} \right)^3} \cr & = 54 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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