If n arithmetic means are inserted between a and 100 such that the ratio of the first mean to the last mean is 1 : 7 and a + n = 33, then the value of n is :
Solution
<p>a, A<sub>1</sub>, A<sub>2</sub> ........... A<sub>n</sub>, 100</p>
<p>Let d be the common difference of above A.P. then</p>
<p>${{a + d} \over {100 - d}} = {1 \over 7}$</p>
<p>$\Rightarrow 7a + 8d = 100$ ...... (i)</p>
<p>and $a + n = 33$ ..... (ii)</p>
<p>and $100 = a + (n + 1)d$</p>
<p>$\Rightarrow 100 = a + (34 - a){{(100 - 7a)} \over 8}$</p>
<p>$\Rightarrow 800 = 8a + 7{a^2} - 338a + 3400$</p>
<p>$\Rightarrow 7{a^2} - 330a + 2600 = 0$</p>
<p>$\Rightarrow a = 10,\,{{260} \over 7},$ but $a \ne {{260} \over 7}$</p>
<p>$\therefore$ $n = 23$</p>
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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