The 10th common term between the series $$3+7+11+\ldots \text { and } 1+6+11+\ldots \text { is }$$ (A) 191 (B) 193 (C) 211 (D) None of these

Step 1: Identify the given arithmetic progressions

We are given two arithmetic progressions (APs). For the first AP, the first term $a_1$ is 3 and the common difference $d_1$ is $7-3=4$. For the second AP, the first term $a_2$ is 1 and the common difference $d_2$ is $6-1=5$.

Step 2: Find the first common term

We list out the terms of both series to find the first common term. The terms of the first series are 3, 7, 11, 15, 19, 23, ... The terms of the second series are 1, 6, 11, 16, 21, 26, ... The first term that appears in both series is 11.

Step 3: Determine the common difference of the common terms

The common terms of two arithmetic progressions also form an arithmetic progression. The common difference of this new AP is the Least Common Multiple (LCM) of the common differences of the original two APs. Here, $d_1 = 4$ and $d_2 = 5$, so the LCM of 4 and 5 is 20.

Step 4: Formulate the AP of common terms

Now we have the first term of the AP of common terms, which is 11, and its common difference, which is 20. So, the AP of common terms is 11, 31, 51, ...

Step 5: Calculate the 10th common term

To find the 10th common term, we use the formula for the $n$ -th term of an AP: $T_n = a + (n-1)D$. Substituting $a = 11$, $n = 10$, and $D = 20$, we get $T_{10} = 11 + (10-1)20 = 11 + 9 \times 20 = 11 + 180 = 191$.

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