Question 1155803
Let {{{ a + 1 }}} be the first of the 39 consecutive numbers
{{{ a + 39 }}} is the 39th of the consecutive numbers
1 through {{{ a }}} are the {{{ a }}} numbers preceding the 39 
consecutive numbers
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The sum of the {{{ 39 + a }}} numbers is:
{{{ ( 39 + a )*( 40 + a ) / 2 }}}
The sum of the 1 through {{{ a }}} numbers is:
{{{ a*( a + 1 ) / 2 }}}
Let the sum of the 39 consecutive numbers be {{{ S }}}
{{{ S =  ( 39 + a )*( 40 + a ) / 2 -  a*( a + 1 ) / 2 }}}
{{{ 2S = 1560 + 40a + 39a + a^2 - a^2 - a }}}
{{{ 2S = 78a + 1560 }}}
{{{ S = 39a + 780 }}}
{{{ 1170 = 39a + 780 }}}
{{{ 39a = 390 }}}
{{{ a = 10 }}}
and
{{{ a + 1 = 11 }}}
{{{ 11 + 38 = 49 }}}
49 is the largest number
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check:
{{{ S =  ( 39 + a )*( 40 + a ) / 2 -  a*( a + 1 ) / 2 }}}
{{{ S =  ( 39 + 10 )*( 40 + 10 ) / 2 -  10*( 10 + 1 ) / 2 }}}
{{{ S = (49*50)/2 - (10*11)/2 }}}
{{{ S = 2450/2 - 110/2 }}}
{{{ S = 1225 - 55 }}}
{{{ S = 1170 }}}
OK
Definitely get a 2nd opinion if needed