Mathematics - Sequence & Series Question with Solution | TestHub

Given that $a,b,c$ are in AP, we have $2b=a+c$. (1) Given that$a,b,c+1$ are in GP, we have $b^2=a(c+1)$. (2) From (1), substitute$c=2b-a$ into (2):$b^2=a(2b-a+1)\Rightarrow b^2=2ab-a^2+a\Rightarrow a^2-2ab+b^2=a\Rightarrow (a-b{)}^2=a$. (3) Since$a$ is a positive integer, $(a-b{)}^2$ must be a perfect square, which implies $a$ must be a perfect square. Let $a=k^2$ for some positive integer $k$. From (3),$(k^2-b{)}^2=k^2\Rightarrow k^2-b=\pm k$. This gives two cases for $b$: Case 1:$k^2-b=k\Rightarrow b=k^2-k=k(k-1)$. Then$c=2b-a=2k(k-1)-k^2=2k^2-2k-k^2=k^2-2k=k(k-2)$. The sequence is$a=k^2$, $b=k(k-1)$, $c=k(k-2)$. For $a,b,c$ to be distinct positive integers, we must have $k(k-2)>0$ and $k(k-1)\ne k^2$ and $k(k-2)\ne k(k-1)$. This implies $k>2$. The smallest integer value for $k$ is $3$. For$k=3$, $a=3^2=9$, $b=3(2)=6$, $c=3(1)=3$. These are distinct positive integers. $a+b+c=9+6+3=18$. Case 2:$k^2-b=-k\Rightarrow b=k^2+k=k(k+1)$. Then$c=2b-a=2k(k+1)-k^2=2k^2+2k-k^2=k^2+2k=k(k+2)$. The sequence is$a=k^2$, $b=k(k+1)$, $c=k(k+2)$. For $a,b,c$ to be distinct positive integers, we must have $k^2>0$ and $k(k+1)\ne k^2$ and $k(k+2)\ne k(k+1)$. This implies $k\ge 1$. The smallest integer value for $k$ is $1$. For$k=1$, $a=1^2=1$, $b=1(2)=2$, $c=1(3)=3$. These are distinct positive integers. $a+b+c=1+2+3=6$. Comparing the sums from both cases, the smallest possible value of$a+b+c$ is 6.