S₁ = 3, 9, 15, ... 25 terms S₂ = 3, 8, 13, ... 37 terms Number of common terms in S₁, S₂ is equal to

The correct option is 5.

Let us analyze the two given arithmetic progressions (APs) step-by-step to find the number of common terms.

Step 1: Understand the first sequence, $S_1$
The first sequence is: $S_1=3,9,15,...$ up to $25$ terms.
For $S_1$:
First term, $a_1=3$
Common difference, $d_1=9-3=6$
Number of terms, $n_1=25$

The general term ($r$-th term) of $S_1$ is given by:
$T_r=a_1+(r-1)d_1$

The last term ($25$-th term) of $S_1$ is:
$L_1=3+(25-1)\times 6=3+24\times 6=3+144=147$

Step 2: Understand the second sequence, $S_2$
The second sequence is: $S_2=3,8,13,...$ up to $37$ terms.
For $S_2$:
First term, $a_2=3$
Common difference, $d_2=8-3=5$
Number of terms, $n_2=37$

The last term ($37$-th term) of $S_2$ is:
$L_2=3+(37-1)\times 5=3+36\times 5=3+180=183$

Step 3: Define the sequence of common terms
Let us find the first common term between $S_1$ and $S_2$ by writing out the initial terms:
$S_1=3,9,15,21,27,33,...$

Clearly, the first common term is $a=3$.

The common terms will also form an arithmetic progression. The common difference of this new AP of common terms is the Least Common Multiple (LCM) of the common differences of $S_1$ and $S_2$:
$d=\text{LCM}(d_1,d_2)=\text{LCM}(6,5)=30$

Thus, the sequence of common terms is:
$3,33,63,93,123,153,...$

Step 4: Find the maximum limit for common terms
Since the common terms must belong to both sequences, they cannot exceed the last term of the shorter sequence. The last term of $S_1$ is $147$ and the last term of $S_2$ is $183$.
Therefore, any common term $t_k$ must satisfy:
$t_k\le \text{min}(147,183)=147$

Step 5: Calculate the number of common terms
Let $k$ be the number of common terms. The $k$-th term of the common sequence is:
$t_k=a+(k-1)d=3+(k-1)\times 30$

Applying the inequality constraint:
$3+(k-1)\times 30\le 147$

Subtracting $3$ from both sides:
$(k-1)\times 30\le 144$

Dividing by $30$:
$k-1\le \frac{144}{30}$
$k-1\le 4.8$
$k\le 5.8$

Since $k$ must be an integer, the maximum possible value for $k$ is $5$.

The $5$ common terms are explicitly:
$3,33,63,93,123$ (all of which are less than or equal to $147$).

Consequently, the number of common terms is $5$.