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

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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 ,

$(\large\frac{a}{d}+n)^2=(\large\frac{a}{d}+m)(\large\frac{a}{d}+r)$

$(\large\frac{a}{d})^2+n^2+2n\;(\large\frac{a}{d})=(\large\frac{a}{d})^2+(\large\frac{a}{d})(m+r)+mr$

$n^2=(\large\frac{a}{d})(m+r-2n)+mr$

$(\large\frac{a}{d})=\large\frac{n^2-mr}{m+r-2n}$

$=\large\frac{n^2-\large\frac{(m+r)n}{2}}{m+r-2n}$

$=\large\frac{\large\frac{n}{2}\;[2n-(m+r)]}{-(2n-(m+r))}$

$=\large\frac{-n}{2}\;.$

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