Official solution
a1 + a7 + a16 = 40
\[{a_1} + \left( {{a_1} + 6d} \right) + ({a_1} + 15d) = 40\]
\[ \Rightarrow 3{a_1} + 21d = 40\]
\[ \Rightarrow {a_1} + 7d = {{40} \over 3}\]
\[ \Rightarrow {a_1} + {a_2}....... + {a_{15}} = {{15} \over 2}[{a_1} + {a_{15}}]\]
\[ \Rightarrow {{15} \over 2}[{a_1} + {a_1} + 14d] \Rightarrow 15({a_1} + 7d) = 15 \times {{40} \over 3} = 200\]