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There are (4n + 1) terms in a certain sequence of which the first (2n + 1) terms form an A.P. of common difference 2 and the last (2n + 1) terms are in G.P. of common ratio 1/2 If the middle term of both A.P. and G.P. are the same, then find the mid-term of this sequence.
Asked by vishnuramrs07 | 30 May, 2023, 21:05: PM
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The space terms space will space be colon straight a comma space straight a plus 2 comma space straight a plus 4 comma space... comma space straight a plus 4 straight n comma space fraction numerator straight a plus 4 straight n over denominator 2 end fraction comma space fraction numerator straight a plus 4 straight n over denominator 2 squared end fraction comma space fraction numerator straight a plus 4 straight n over denominator 2 cubed end fraction comma space... space fraction numerator straight a plus 4 straight n over denominator 2 to the power of 2 straight n end exponent end fraction As space the space middle space terms space of space straight A. straight P. space and space straight G. straight P. space are space equal rightwards double arrow straight a plus 2 straight n equals fraction numerator straight a plus 4 straight n over denominator 2 to the power of straight n end fraction rightwards double arrow straight a open parentheses 1 minus 1 over 2 to the power of straight n close parentheses equals fraction numerator 4 straight n over denominator 2 to the power of straight n end fraction minus 2 straight n rightwards double arrow straight a equals fraction numerator 4 straight n minus 2 to the power of straight n times 2 straight n over denominator 2 to the power of straight n minus 1 end fraction So comma space the space middle space term space of space the space sequence space is straight a plus 4 straight n equals fraction numerator 4 straight n minus 2 to the power of straight n times 2 straight n over denominator 2 to the power of straight n minus 1 end fraction plus 4 straight n equals fraction numerator 4 straight n minus 2 to the power of straight n times 2 straight n plus 4 straight n times 2 to the power of straight n minus 4 straight n over denominator 2 to the power of straight n minus 1 end fraction equals fraction numerator 2 to the power of straight n times 2 straight n over denominator 2 to the power of straight n minus 1 end fraction
Answered by Renu Varma | 31 May, 2023, 11:27: AM
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