Question

find three consecutive numbers such that the sum of second and the third number exceeds the first by 14

Answer 1

Let the three consecutive numbers be x, ( x + 1 ) and ( x + 2 ).

Since the sum of second and third consecutive number exceeds the first by 14 we can represent it in the form of an equation as follows:

( x + 1 ) + ( x + 2 ) = x + 14
=> 2x + 3 = x + 14
=> x = 11

Consecutive numbers = 11, 12 and 13

Answer 2

hey
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let the first no be x
second be x+1
third be x+2

a/q its given sum of second and third exceeds the first no by 14

(x+1)+(x+2) =x+14
2x+3=x+14
2x-x=14-3
x=11

•first no=11
•second=11+1=12
•third no=11+2=13

hence, numbers are 11,12,13

hope helped
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