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8. Sequences and Series
hard
ધારો કે $a_1, a_2, \ldots, a_n$ સમાંતર શ્રેણીમાં છ. જો $a_5=2 a_7$ અને $a_{11}=18$ હોય, તો $12\left(\frac{1}{\sqrt{a_{10}}+\sqrt{a_{11}}}+\frac{1}{\sqrt{a_{11}}+\sqrt{a_{12}}}+\ldots . \cdot \frac{1}{\sqrt{a_{17}}+\sqrt{a_{18}}}\right)=................$
A
$8$
B
$6$
C
$3$
D
$12$
(JEE MAIN-2023)
Solution
$2 a_7=a_s(\text { given })$
$2\left(a_1+6 d\right)=a_1+4 d$
$a_1+8 d=0$
$a_1+10 d=18$
$\text { By }(1) \text { and }(2) \text { we get } a_1=-72, d=9$
$a_{18}=a_1+17 d=-72+153=81$
$a_{10}=a_1+9 d=9$
$12\left(\frac{\sqrt{a_{11}}-\sqrt{a_{10}}}{d}+\frac{\sqrt{a_{12}}-\sqrt{a_{11}}}{d}+\ldots . . \frac{\sqrt{a_{18}}-\sqrt{a_{17}}}{d}\right)$
$12\left(\frac{\sqrt{a_{18}}-\sqrt{a_{10}}}{d}\right)=\frac{12(9-3)}{9}=\frac{12 \times 6}{6}=8$
Standard 11
Mathematics
