UPSC CDS 1 2021 Math Paper

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