Question 1124696

(1) {{{ a[1] + (a[1]+d) + (a[1]+2d) = 36 }}}
(2) {{{ (a[1]+3d) + (a[1]+4d) + (a[1]+5d) = 72 }}}

(1) ==>  {{{3a[1] + 3d = 36 }}}
(2) ==> {{{3a[1] + 12d = 72 }}}

Subtract (1) from (2):
              {{{ 9d = 36 }}} —>   {{{ highlight( d = 4 ) }}}

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Another approach:  
 In the first group of three, the average value is 36/3 = 12  (= middle value, {{{a[2]}}} )
 In the 2nd group of three, the average value is 72/3 = 24  (= middle value, {{{a[5]}}} )

Since {{{ a[5] = a[2] + 3d }}}:  24 = 12 + 3d —> 3d = 12 —> d = 4