Let the two alternate numbers be x and (x+2) Now, we obtain the sum of the reciprocals as: (1∕x)+[1∕(x+2)]=7∕24 ⇒(x+x+2)∕(x2+2x)=7∕24 ⇒24×(2x+2)=7×(x2+2x) ⇒48x+48=7x2+14x ⇒48x−14x=7x2−48 ⇒34x=7x2−48 ⇒7x2−34x−48=0 ⇒(x−6)[x+(8∕7)]=0 ⇒x=6,−(8∕7) We are given the natural numbers only. So, no negative number will be allowed for x. Hence, thenext natural number =6+2=8 ∴ The sum of the numbers =6+8=14