Let A = \ 1,a_1,a_2, ... ,a_18,77 \ be a set of integers with $1

a_1, a_2, a_3, ... , a_18, 77 are in AP i.e. 1, 5, 9, 13, ..., 77. Hence a_1+a_2+a_3+ ... +a_18 =5+9+13+ ... 18 terms =702

Let A = \{ 1,a_{1},a_{2},,a_{18},77 \} be a set of in... | Sequence & Series (Mathematics)

Let

A = \left\{ 1,a_{1},a_{2},\ldots,a_{18},77 \right\}

be a set of integers with

1 < a_{1} < a_{2} < \ldots < a_{18} < 77

. Let the set

+ A = \{ x + y

x,y \in A\}

contain exactly 39 elements. Then, the value of

a_{1} + a_{2} + \ldots + a_{18}

is equal to