Key Concepts and Formulas

• Arithmetic Progression (AP): A sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the common difference ($d$). The general term is $a_n = a + (n-1)d$, where $a$ is the first term.
• Common Terms of Multiple APs: The common terms of several APs themselves form an AP.
  • The common difference of this new AP is the Least Common Multiple (LCM) of the common differences of the original APs.
  • The first term of this new AP is the smallest term common to all original APs.
  • The last term of this new AP is the largest term common to all original APs, which must be less than or equal to the minimum of the last terms of the individual APs.
• Sum of an AP: The sum of an AP with $N$ terms, first term $a$, and last term $L$ is $S_N = \frac{N}{2}(a + L)$.

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Step-by-Step Solution

Step 1: Analyze Each Arithmetic Progression (AP) We first identify the first term ($a$), common difference ($d$), and last term ($L$) for each of the three given APs. This is crucial for determining the properties of the AP formed by their common terms.

• AP 1: $3, 7, 11, 15, \ldots, 399$

  • First term, $a_1 = 3$.
  • Common difference, $d_1 = 7 - 3 = 4$.
  • Last term, $L_1 = 399$.

• AP 2: $2, 5, 8, 11, \ldots, 359$

  • First term, $a_2 = 2$.
  • Common difference, $d_2 = 5 - 2 = 3$.
  • Last term, $L_2 = 359$.

• AP 3: $2, 7, 12, 17, \ldots, 197$

  • First term, $a_3 = 2$.
  • Common difference, $d_3 = 7 - 2 = 5$.
  • Last term, $L_3 = 197$.

Step 2: Determine the Common Difference of the AP of Common Terms A term common to all three APs must advance in steps that are multiples of each individual common difference. Therefore, the common difference of the AP formed by the common terms will be the LCM of the common differences of the original APs.

The common differences are $d_1=4$, $d_2=3$, and $d_3=5$. Let $D$ be the common difference of the common terms. $D = \text{LCM}(d_1, d_2, d_3) = \text{LCM}(4, 3, 5)$ Since $4, 3, 5$ are pairwise coprime, their LCM is their product: $D = 4 \times 3 \times 5 = 60$

Step 3: Find the First Common Term To find the first term common to all three APs, we need to find the smallest number that belongs to all three sequences. We can do this by observing the pattern of each AP or by solving a system of congruences. The general form of terms in each AP is:

• AP 1: $3 + 4k_1$
• AP 2: $2 + 3k_2$
• AP 3: $2 + 5k_3$

We are looking for the smallest number $x$ such that: $x \equiv 3 \pmod{4}$ $x \equiv 2 \pmod{3}$ $x \equiv 2 \pmod{5}$

From $x \equiv 2 \pmod{3}$ and $x \equiv 2 \pmod{5}$, since 3 and 5 are coprime, we have $x \equiv 2 \pmod{\text{LCM}(3,5)}$, which means $x \equiv 2 \pmod{15}$. So, $x$ can be written in the form $15k + 2$. Now we substitute this into the first congruence: $15k + 2 \equiv 3 \pmod{4}$ $15k \equiv 1 \pmod{4}$ Since $15 \equiv 3 \pmod{4}$, we have: $3k \equiv 1 \pmod{4}$ Multiplying by 3 (which is the inverse of 3 mod 4, since $3 \times 3 = 9 \equiv 1 \pmod{4}$): $9k \equiv 3 \pmod{4}$ $k \equiv 3 \pmod{4}$ So, $k$ can be written as $4m + 3$. Substituting this back into $x = 15k + 2$: $x = 15(4m + 3) + 2$ $x = 60m + 45 + 2$ $x = 60m + 47$ The smallest positive value for $x$ is when $m=0$, which gives $x=47$. Thus, the first common term, $A$, is $47$.

Step 4: Determine the Last Common Term The common terms must exist within the range of all three APs. Therefore, the last common term cannot exceed the smallest of the last terms of the individual APs.

The last terms are $L_1=399$, $L_2=359$, and $L_3=197$. The maximum possible value for a common term is $\min(L_1, L_2, L_3) = \min(399, 359, 197) = 197$.

The AP of common terms starts with $A=47$ and has a common difference $D=60$. The terms are of the form $47 + (N-1)60$. We need to find the largest term of this form that is less than or equal to $197$. The terms are:

• $47$ (for $N=1$)
• $47 + 60 = 107$ (for $N=2$)
• $47 + 2 \times 60 = 47 + 120 = 167$ (for $N=3$) The next term would be $47 + 3 \times 60 = 47 + 180 = 227$, which is greater than $197$. Therefore, the last common term, $L'$, is $167$.

Step 5: Calculate the Number of Common Terms The common terms form an AP with first term $A=47$, common difference $D=60$, and last term $L'=167$. We can find the number of terms ($N$) using the formula $L' = A + (N-1)D$.

$167 = 47 + (N-1)60$ Subtract $47$ from both sides: $120 = (N-1)60$ Divide by $60$: $2 = N-1$ $N = 3$ There are $3$ common terms.

Step 6: Calculate the Sum of the Common Terms Now we calculate the sum of these $N=3$ common terms using the formula $S_N = \frac{N}{2}(A + L')$.

$S_3 = \frac{3}{2}(47 + 167)$ $S_3 = \frac{3}{2}(214)$ $S_3 = 3 \times 107$ $S_3 = 321$

The sum of the common terms is $321$.

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Common Mistakes & Tips

• Incorrect LCM: Ensure the LCM of common differences is calculated correctly, especially for numbers that are not pairwise coprime.
• Forgetting the Upper Bound: Always find the minimum of the last terms of the given APs to correctly determine the last common term.
• Confusing Number of Terms with Sum: The question asks for the sum of common terms. Be sure to perform the final summation step. In this case, there are 3 common terms, but their sum is 321.

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Summary To find the sum of common terms of three APs, we first determine the properties of each AP. The common terms themselves form a new AP whose common difference is the LCM of the individual common differences. The first term of this new AP is the smallest term common to all original APs, and its last term is the largest common term that does not exceed the minimum of the last terms of the original APs. We then calculate the number of terms in this common AP and use the sum formula to find the total sum. The common terms are $47, 107, 167$. There are $3$ such terms, and their sum is $321$.

The final answer is $\boxed{321}$.