Mathematics - Sequence & Series Question with Solution | TestHub

Let the A.P. be $a_1,a_1+d,\dots ,a_1+(n-1)d$. Given $a_1=1$ and $a_k$ are positive integers, so $d$ must be a non-negative integer. The sum $S_n=\frac{n}{2}(2a_1+(n-1)d)=\frac{n}{2}(2+(n-1)d)$. Given $S_n=m^2$ for some integer $m$. Also, $a_n=1+(n-1)d<100$.

Case 1: $n$ is odd. Let $n=2k+1$. $S_n=(2k+1)(1+kd)=m^2$. For this to be a square, $1+kd=(2k+1)j^2$ for some integer $j\ge 1$. Then $d=\frac{(2k+1)j^2-1}{k}$. The condition $a_n<100$ means $1+(2k)d<100⟹2kd<99$. Substituting $d$: $2k\left(\frac{(2k+1)j^2-1}{k}\right)<99⟹2((2k+1)j^2-1)<99⟹2(2k+1)j^2<101$. Since $j\ge 1$, $2(2k+1)<101⟹2k+1<50.5$. So $n=2k+1\le 49$. For $n=49$ ($k=24$), $98j^2<101⟹j=1$. Then $d=\frac{49(1{)}^2-1}{24}=2$. $a_{49}=1+(49-1)2=97<100$. So $n=49$ is possible.

Case 2: $n$ is even. Let $n=2k$. $S_n=k(2+(2k-1)d)=m^2$. The condition $a_n<100$ means $1+(2k-1)d<100$. To maximize $n$, let $k=j^2$ for some integer $j$. Then $S_n=j^2(2+(2j^2-1)d)=m^2$, so $2+(2j^2-1)d$ must be a perfect square, say $L^2$. Thus $d=\frac{L^2-2}{2j^2-1}$. The condition $a_n<100$ means $1+(L^2-2)<100⟹L^2<101$. So $L_{max}=10$. If $L=10$, $a_n=1+(100-2)=99<100$. Then $d=\frac{100-2}{2j^2-1}=\frac{98}{2j^2-1}$. For $d$ to be an integer, $2j^2-1$ must be an odd divisor of 98. Possible values for $2j^2-1$ are $1,7,49$. If $2j^2-1=49⟹2j^2=50⟹j^2=25⟹j=5$. Then $n=2k=2j^2=50$. $d=98/49=2$. $a_{50}=1+(50-1)2=99<100$. So $n=50$ is possible.

Comparing the maximum values from both cases, $n=49$ (odd) and $n=50$ (even), the maximum possible value of $n$ is 50.