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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
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If $(m+n)^{th}$, $(n+1)^{th}$, $(r+1)^{th}$ term of an AP are in GP and $m, n, r$ are in HP, then ratio of first term of AP to common difference is

$(a)\;n\qquad(b)\;\large\frac{n}{2}\qquad(c)\;\large\frac{n}{3}\qquad(d)\;\large\frac{-n}{2}$

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Answer : (d) $\;\large\frac{-n}{2}$
Explanation : $a_{m+1}\;,a_{n+1}\;a_{r+1}\;$ are in GP
$(a_{n+1})^2=a_{m+1}\;.a_{r+1}$
$(a+nd)^2=(a+md)(a+rd)$
m , n , r are in HP
$n=\large\frac{2mr}{(m+r)}$
So ,
$(\large\frac{a}{d}+n)^2=(\large\frac{a}{d}+m)(\large\frac{a}{d}+r)$
$(\large\frac{a}{d})^2+n^2+2n\;(\large\frac{a}{d})=(\large\frac{a}{d})^2+(\large\frac{a}{d})(m+r)+mr$
$n^2=(\large\frac{a}{d})(m+r-2n)+mr$
$(\large\frac{a}{d})=\large\frac{n^2-mr}{m+r-2n}$
$=\large\frac{n^2-\large\frac{(m+r)n}{2}}{m+r-2n}$
$=\large\frac{\large\frac{n}{2}\;[2n-(m+r)]}{-(2n-(m+r))}$
$=\large\frac{-n}{2}\;.$
answered Jan 24, 2014 by yamini.v
 

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