find three consecutive odd numbers whose run is 147 (2x+1) (2x+3) (2x+5)
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Heya..!!!
given that:-
Sum of three consecutive terms = 147
(2x + 1) + (2x + 3) + (2x + 5) = 147
=> 6x + 9 = 147
=> 6x = 147 - 9
=> X = 138/2
=> X = 69
NOW the no. Be,
(2x + 1) = (2×69 + 1) = 138 + 1 = 139
(2x + 3) = (2×69 + 3) = 138 + 3 = 141
(2x + 5) = (2×69 + 5) = 138 + 5 = 143
given that:-
Sum of three consecutive terms = 147
(2x + 1) + (2x + 3) + (2x + 5) = 147
=> 6x + 9 = 147
=> 6x = 147 - 9
=> X = 138/2
=> X = 69
NOW the no. Be,
(2x + 1) = (2×69 + 1) = 138 + 1 = 139
(2x + 3) = (2×69 + 3) = 138 + 3 = 141
(2x + 5) = (2×69 + 5) = 138 + 5 = 143
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