Given that, $ a_{1} + a_{2} + a_{3} + \dots + a_{n} = 3(2^{n+1} – 2) $
Now, put $ n=11,$ we get
$ a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}+a_{7}+a_{8}+a_{9}+a_{10}+a_{11} = 3(2^{12}-2) $
$ = 3 (4096-2) $
$ = 3 \times 4094 $
$ = 12282 \quad \longrightarrow (1) $
And put $ n=10,$ we get
$ a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}+a_{7}+a_{8}+a_{9}+a_{10} = 3(2^{11}-2) $
$ = 3 (2048 – 2) $
$ = 3 \times 2046 $
$ = 6138 \quad \longrightarrow (2) $
From the equation $(1),$ subtract the equation $(2),$ we get
$ a_{11} = 12282 – 6138 $
$ \boxed{a_{11} = 6144} $
Correct Answer $: 6144$