NIMCET 2017 Question Paper with solutions

Consider, three positive number a−d,a,a+d are in AP
Three positive number whose sum is 21 .
⇒a−d+a+a+d=21
⇒a=7
Hence, three positive number becomes 7−d,7,7+d.
It is given, If 2,2,14 are added to them respectively then resulting number are in geometric progression.
⇒9−d,9,21+d are in GP.
⇒‌

9

9−d

=‌

21+d

9

⇒81=(21+d)(9−d)
⇒d2+12d−108=0
⇒(d+18)(d−6)=0
⇒d=−18 and d=6
When d=−18, the three numbers are 25,7,−11
When d=6, the three numbers are 1,7,13