Which term of the G.P. :
−10,53,−56,........ is −572?-10, \dfrac{5}{\sqrt{3}}, -\dfrac{5}{6}, …….. \text{ is } -\dfrac{5}{72}?−10,35,−65,…….. is −725?
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Common ratio (r) = 53−10=−5103=−123\dfrac{\dfrac{5}{\sqrt{3}}}{-10} = -\dfrac{5}{10\sqrt{3}} = -\dfrac{1}{2\sqrt{3}}−1035=−1035=−231.
Let nth term of G.P. be −572-\dfrac{5}{72}−725.
∴arn−1=−572⇒−10×(−123)n−1=−572⇒(−123)n−1=−572×−110⇒(−123)n−1=1144⇒(−123)n−1=(−123)4⇒n−1=4⇒n=5.\therefore ar^{n - 1} = -\dfrac{5}{72} \\[1em] \Rightarrow -10 \times \Big(-\dfrac{1}{2\sqrt{3}}\Big)^{n - 1} = -\dfrac{5}{72} \\[1em] \Rightarrow \Big(-\dfrac{1}{2\sqrt{3}}\Big)^{n - 1} = -\dfrac{5}{72} \times -\dfrac{1}{10} \\[1em] \Rightarrow \Big(-\dfrac{1}{2\sqrt{3}}\Big)^{n - 1} = \dfrac{1}{144} \\[1em] \Rightarrow \Big(-\dfrac{1}{2\sqrt{3}}\Big)^{n - 1} = \Big(-\dfrac{1}{2\sqrt{3}}\Big)^4 \\[1em] \Rightarrow n - 1 = 4 \\[1em] \Rightarrow n = 5.∴arn−1=−725⇒−10×(−231)n−1=−725⇒(−231)n−1=−725×−101⇒(−231)n−1=1441⇒(−231)n−1=(−231)4⇒n−1=4⇒n=5.
Hence, 5th term of the G.P. is −572.-\dfrac{5}{72}.−725.
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Find the seventh term of the G.P. :
3+1,1,3−12,\sqrt{3} + 1, 1, \dfrac{\sqrt{3} - 1}{2},3+1,1,23−1, ………..
Find the next two terms of the series :
2 - 6 + 18 - 54 ………….
The fifth term of a G.P. is 81 and its second term is 24. Find the geometric progression.
Fourth and seventh terms of a G.P. are 118 and −1486\dfrac{1}{18} \text{ and } -\dfrac{1}{486}181 and −4861 respectively. Find the G.P.
