For a natural number n, let ${\alpha _n} = {19^n} - {12^n}$. Then, the value of ${{31{\alpha _9} - {\alpha _{10}}} \over {57{\alpha _8}}}$ is ___________.
Answer (integer)
4
Solution
<p>${\alpha _n} = {19^n} - {12^n}$</p>
<p>Let equation of roots 12 & 19 i.e.</p>
<p>${x^2} - 31x + 228 = 0$</p>
<p>$\Rightarrow (31 - x) = {{228} \over x}$ (where x can be 19 or 12)</p>
<p>$\therefore$ $${{31{\alpha _9} - {\alpha _{10}}} \over {57{\alpha _8}}} = {{31({{19}^9} - {{12}^9}) - ({{19}^{10}} - {{12}^{10}})} \over {57({{19}^8} - {{12}^8})}}$$</p>
<p>$= {{{{19}^9}(31 - 19) - {{12}^9}(31 - 12)} \over {57({{19}^8} - {{12}^8})}}$</p>
<p>$= {{228({{19}^8} - {{12}^8})} \over {57({{19}^8} - {{12}^8})}} = 4$.</p>
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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