In general, there are two classification for abelian groups. One is isomorphic to a direct sum of cyclic groups of order equal to a prime power, and the other is isomorphic to a direct sum of cyclic groups of order equal to a positive integer.
Notice that $4900=2^2\times 5^2\times 7^2$.
Let $G$ be isomorphic to a direct sum of cyclic groups of order equal to a prime power. In this case, we denote all isomorphism classes by arrays $(49,25,4)$,$(49,25,2,2)$,$(49,5,5,4)$,
$(49,5,5,2,2)$,$(7,7,25,4)$,$(7,7,25,2,2)$,$(7,7,5,5,4)$,$(7,7,5,5,2,2)$. That is
$$\begin{align}G&\cong \mathbb{Z}_{49}\times \mathbb{Z}_{25}\times \mathbb{Z}_{4}\\
G&\cong \mathbb{Z}_{49}\times \mathbb{Z}_{25}\times \mathbb{Z}_{2}\times \mathbb{Z}_{2}\\
G&\cong \mathbb{Z}_{49}\times \mathbb{Z}_{5}\times \mathbb{Z}_{5}\times \mathbb{Z}_{4}\\
G&\cong \mathbb{Z}_{49}\times \mathbb{Z}_{5}\times \mathbb{Z}_{5}\times \mathbb{Z}_{2}\times \mathbb{Z}_{2}\\
G&\cong \mathbb{Z}_{7}\times \mathbb{Z}_{7}\times \mathbb{Z}_{25}\times \mathbb{Z}_{4}\\
G&\cong \mathbb{Z}_{7}\times \mathbb{Z}_{7}\times \mathbb{Z}_{25}\times \mathbb{Z}_{2}\times \mathbb{Z}_{2}\\
G&\cong \mathbb{Z}_{7}\times \mathbb{Z}_{7}\times \mathbb{Z}_{5}\times \mathbb{Z}_{5}\times \mathbb{Z}_{4}\\
G&\cong \mathbb{Z}_{7}\times \mathbb{Z}_{7}\times \mathbb{Z}_{5}\times \mathbb{Z}_{5}\times \mathbb{Z}_{2}\times \mathbb{Z}_{2}\end{align}
$$
If we apply the fundamental theorem of finite abelian groups in the form "there are $d_i$, s.t. $d_i\mid d_{i+1}$ and $G\cong \mathbb{Z}_{d_1}\times \mathbb{Z}_{d_2}\times \cdots \times \mathbb{Z}_{d_s}$", then we rearrange the above results using $\mathbb{Z}_{nm}\cong \mathbb{Z}_m\times \mathbb{Z}_n$, where $\gcd(n,m)=1$. Thus we get
$$\begin{align}G&\cong \mathbb{Z}_{4900}\\
G&\cong \mathbb{Z}_{2}\times \mathbb{Z}_{2950}\\
G&\cong \mathbb{Z}_{5}\times \mathbb{Z}_{980}\\
G&\cong \mathbb{Z}_{10}\times \mathbb{Z}_{490}\\
G&\cong \mathbb{Z}_{7}\times \mathbb{Z}_{700}\\
G&\cong \mathbb{Z}_{14}\times \mathbb{Z}_{350}\\
G&\cong \mathbb{Z}_{35}\times \mathbb{Z}_{140}\\
G&\cong \mathbb{Z}_{70}\times \mathbb{Z}_{70}\end{align}
$$
Best Answer
Your idea is correct. At the beginning of the proof is stated that $G\cong\Bbb Z_{d_1}\times...\Bbb Z_{d_r}$, where $d_i$ is a power of a prime for every $i$. Because of this, $G$ is cyclic if and only if $(d_i,d_j)=1$ for every $i \neq j$ (You can prove this as an exercise). Now consider that if for some distinct $i,j$ the numbers $d_i$ and $d_j$ share a prime factor, then $m=lcm( d_1, d_2, \dots d_r)$ is strictly less than $ d_1 d_2 \dots d_r$. Therefore, if these two numbers are equal, $(d_i,d_j)=1$ for every $i \neq j$ .