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.

To find the number of common terms in the two arithmetic progressions (APs), we first analyze each sequence separately.

1. Analysis of the first sequence S₁:
The terms are: 3, 9, 15, ... up to 25 terms.
First term, $a_1=3$
Common difference, $d_1=9-3=6$
Number of terms, $n_1=25$
The last term of S₁ is given by:
$T_{25}=a_1+(25-1)d_1$
$T_{25}=3+24\times 6=3+144=147$

2. Analysis of the second sequence S₂:
The terms are: 3, 8, 13, ... up to 37 terms.
First term, $a_2=3$
Common difference, $d_2=8-3=5$
Number of terms, $n_2=37$
The last term of S₂ is given by:
${\mathrm{T\text{'}}}_{37}=a_2+(37-1)d_2$
${\mathrm{T\text{'}}}_{37}=3+36\times 5=3+180=183$

3. Determining the common terms:
The first common term of both sequences is:
$A=3$
The common difference of the common terms series is the Least Common Multiple (LCM) of the common differences of both APs:
$d=\text{LCM}(d_1,d_2)=\text{LCM}(6,5)=30$
Therefore, the common terms form a new arithmetic progression starting with 3 and having a common difference of 30:
3, 33, 63, 93, ...

Let the number of common terms be $k$. The last common term must not exceed the smaller of the last terms of both individual series. That is:
$T_{\mathrm{common}}\le min(147,183)=147$

Using the formula for the general term of an AP:
$A+(k-1)d\le 147$
$3+(k-1)\times 30\le 147$
$(k-1)\times 30\le 144$
$k-1\le \frac{144}{30}$
$k-1\le 4.8$
$k\le 5.8$

Since the number of terms $k$ must be a positive integer, the maximum value of $k$ is 5.

Thus, the number of common terms in S₁ and S₂ is 5.