Progressions Part 4

Algebra Level pending

Find the 1 0 t h 10^{th} term of the following series,

2 , 1 2 , 1 2 2 . . . . . . . . . \huge \sqrt{2}, \dfrac{1}{\sqrt{2}}, \dfrac{1}{2\sqrt{2}} . . . . .. . . .

b : b: 2 256 \dfrac{2}{\sqrt{256}} a : a: 1 256 2 \dfrac{1}{256\sqrt{2}} . c : c: 1 2 512 \dfrac{1}{2\sqrt{512}} d : d: 1 2 128 \dfrac{1}{2\sqrt{128}}

Sravanth C. by Sravanth C. , 21, India

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1 solution

Sravanth C.
May 22, 2015

According to the question, first term a = 2 a=\sqrt{2} and the common ratio r = 1 2 2 = 1 2 r=\dfrac{\dfrac{1}{\sqrt{2}}}{\sqrt{2}}=\dfrac{1}{2}

We know that, t n = a r n − 1 t_{n}=ar^{n-1}

Therefore, t 10 = 2 × 1 2 9 = 2 × 1 512 = 2 × 2 512 × 2 = 2 512 2 = 1 256 2 t_{10}=\sqrt{2}×\dfrac{1}{2^{9}} \\ =\sqrt{2}×\dfrac{1}{512} \\ = \dfrac{\sqrt{2}×\sqrt{2}}{512×\sqrt{2}} \\ =\dfrac{2}{512\sqrt{2}} \\ =\boxed{\dfrac{1}{256\sqrt{2}}} \\

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