Let a_1,a_2,… ,a_n be in A.P. If a_5=2a_7 and a_11=18, then 12( 1 √a_10+√a_11+ 1 √a_11+√a_12+… .. 1 √a_17+√a_18) is equal to .

Let the common difference of an A.P. be $d$d, then

$2{a}_{7}={a}_{5}$2a7​=a5​ (given)

$$\Rightarrow 2({a}_{1}+6d)={a}_{1}+4d$$⇒2(a1​+6d)=a1​+4d

$$\Rightarrow {a}_{1}+8d=0\ldots (1)$$⇒a1​+8d=0…(1)

And,

$${a}_{11}=18$$a11​=18

$$\Rightarrow {a}_{1}+10d=18\ldots (2)$$⇒a1​+10d=18…(2)

Solving $(1)$(1) and $(2)$(2), we get

$${a}_{1}=-72,d=9$$a1​=−72,d=9

So,

$${a}_{18}={a}_{1}+17d=-72+153=81$$a18​=a1​+17d=−72+153=81

$${a}_{10}={a}_{1}+9d=9$$a10​=a1​+9d=9

Now,

$$12(\frac{1}{\sqrt{{a}_{10}}+\sqrt{{a}_{11}}}+\frac{1}{\sqrt{{a}_{11}}+\sqrt{{a}_{12}}}+\ldots ..\frac{1}{\sqrt{{a}_{17}}+\sqrt{{a}_{18}}})$$12(a10​​+a11​​1​+a11​​+a12​​1​+…..a17​​+a18​​1​)

$$=12(\frac{\sqrt{{a}_{11}}-\sqrt{{a}_{10}}}{d}+\frac{\sqrt{{a}_{12}}-\sqrt{{a}_{11}}}{d}+\ldots \ldots \frac{\sqrt{{a}_{18}}-\sqrt{{a}_{17}}}{d})$$=12(da11​​−a10​​​+da12​​−a11​​​+……da18​​−a17​​​)

$$=12(\frac{\sqrt{{a}_{18}}-\sqrt{{a}_{10}}}{d}) =\frac{12(9-3)}{9}=8$$=12(da18​​−a10​​​)=912(9−3)​=8