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माना $\mathrm{a}_1, \mathrm{a}_2, \ldots \ldots, \mathrm{a}_{\mathrm{n}}$ $A.P.$ में हैं। यदि $\mathrm{a}_5=2 \mathrm{a}_7$ तथा $\mathrm{a}_{11}=18$ है, तो $12\left(\frac{1}{\sqrt{a_{10}}+\sqrt{a_{11}}}+\frac{1}{\sqrt{a_{11}}+\sqrt{a_{12}}}+\ldots . \cdot \frac{1}{\sqrt{a_{17}}+\sqrt{a_{18}}}\right)$ बराबर है_________.
$8$
$6$
$3$
$12$
Solution
$2 a_7=a_s(\text { given })$
$2\left(a_1+6 d\right)=a_1+4 d$
$a_1+8 d=0$
$a_1+10 d=18$
$\text { By }(1) \text { and }(2) \text { we get } a_1=-72, d=9$
$a_{18}=a_1+17 d=-72+153=81$
$a_{10}=a_1+9 d=9$
$12\left(\frac{\sqrt{a_{11}}-\sqrt{a_{10}}}{d}+\frac{\sqrt{a_{12}}-\sqrt{a_{11}}}{d}+\ldots . . \frac{\sqrt{a_{18}}-\sqrt{a_{17}}}{d}\right)$
$12\left(\frac{\sqrt{a_{18}}-\sqrt{a_{10}}}{d}\right)=\frac{12(9-3)}{9}=\frac{12 \times 6}{6}=8$
