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Step-by-Step Solution
Step 1: Identify the two arithmetic progressions (APs)
• First AP: 3, 7, 11, 15, …, 403
– First term a_1 = 3
– Common difference d_1 = 4
• Second AP: 2, 5, 8, 11, …, 404
– First term a'_1 = 2
– Common difference d'_1 = 3
Step 2: Express the general terms of each AP
• General term of the first AP (let its n th term be T_n ):
T_n = 3 + (n - 1)\times 4.
• General term of the second AP (let its m th term be T'_m ):
T'_m = 2 + (m - 1)\times 3.
Step 3: Find the condition for common terms
A common term x must satisfy
x = 3 + 4k \quad\text{and}\quad x = 2 + 3m
for some integers k and m . Equating these,
3 + 4k = 2 + 3m
\quad\Longrightarrow\quad
4k - 3m = -1.
Step 4: Determine the first common term
We look for integer solutions (k, m) .
Try k = 2 :
4(2) - 3m = -1
\quad\Longrightarrow\quad
8 - 3m = -1
\quad\Longrightarrow\quad
3m = 9
\quad\Longrightarrow\quad
m = 3.
This results in
x = 3 + 4\times 2 = 11,
which is also 2 + 3\times 3 = 11.
So, 11 is the first common term.
Step 5: Find the common difference among the common terms
The next common terms occur in steps of \mathrm{LCM}(4, 3) = 12.
Hence, the sequence of common terms is itself an AP with
• First common term = 11,
• Common difference = 12.
Step 6: Determine the number of common terms up to the largest term in either AP
The common terms must not exceed the smaller of the last terms 403 and 404.
Let the total number of these common terms be n.
The n th common term is
11 + (n - 1)\times 12.
We require
11 + 12(n - 1) \le 403.
Thus,
12(n - 1) \le 392
\quad\Longrightarrow\quad
n - 1 \le \frac{392}{12}
\quad\Longrightarrow\quad
n - 1 \le 32.66...
\quad\Longrightarrow\quad
n \le 33.
Hence, there are 33 common terms.
The largest common term is
11 + 12\times(33 - 1) = 11 + 384 = 395.
Step 7: Calculate the sum of these common terms
The 33 common terms form an AP with first term a = 11 and common difference d = 12.
The sum of n terms of an AP is given by
S = \frac{n}{2}\,\bigl[2a + (n-1)d \bigr].
Substituting n=33, a=11, and d=12, we get:
S = \frac{33}{2} \bigl[ 2 \times 11 + (33 - 1)\times 12 \bigr].
Inside the bracket:
2 \times 11 = 22, \quad (33 - 1)\times 12 = 32\times 12 = 384,
so
2a + (n-1)d = 22 + 384 = 406.
Therefore,
S = \frac{33}{2}\times 406 = 33 \times 203 = 6699.
Final Answer
The sum of the common terms in the two given arithmetic progressions is 6699.
