Let 3, 7, 11, 15, ...., 403 and 2, 5, 8, 11, . . ., 404 be two arithmetic progressions. Then the sum, of the common terms in them, is equal to _________.
Correct Answer: 6699
Approach Solution - 1
The first arithmetic progression (AP) is:
3, 7, 11, 15, ..., 403
The second arithmetic progression (AP) is:
2, 5, 8, 11, ..., 404
To find the common terms, we first find the least common multiple (LCM) of the common differences of both progressions:
\[ \text{LCM}(4, 3) = 12 \]
The sequence of common terms is:
11, 23, 35, ..., 403
This is an AP with first term \(a = 11\) and common difference \(d = 12\). We need to find the number of terms (\(n\)) in this AP such that the last term is 403:
\[ 403 = 11 + (n - 1) \times 12 \]
\[ 392 = (n - 1) \times 12 \implies n - 1 = \frac{392}{12} = 32 \implies n = 33 \]
The sum of the common terms is given by:
\[ S_n = \frac{n}{2} [2a + (n - 1) \times d] \]
Substituting the values:
\[ S_{33} = \frac{33}{2} [2 \times 11 + (33 - 1) \times 12] \]
\[ = \frac{33}{2} [22 + 32 \times 12] \]
\[ = \frac{33}{2} \times 406 = 6699 \]
Approach Solution -2
We have two arithmetic progressions:
AP₁: 3, 7, 11, 15, ..., 403
AP₂: 2, 5, 8, 11, ..., 404
We need to find the sum of the common terms in both sequences.
Concept Used:
The common terms of two arithmetic progressions form another arithmetic progression. If AP₁ has first term \(a_1\), common difference \(d_1\), and last term \(l_1\), and AP₂ has first term \(a_2\), common difference \(d_2\), and last term \(l_2\), then the common terms form an AP with:
- First term = first number common to both sequences
- Common difference = LCM(\(d_1, d_2\))
- Last term = last number common to both sequences
Step-by-Step Solution:
Step 1: Identify parameters of both APs.
AP₁: First term \(a_1 = 3\), common difference \(d_1 = 4\), last term \(l_1 = 403\)
Number of terms: \(n_1 = \frac{403 - 3}{4} + 1 = \frac{400}{4} + 1 = 101\)
AP₂: First term \(a_2 = 2\), common difference \(d_2 = 3\), last term \(l_2 = 404\)
Number of terms: \(n_2 = \frac{404 - 2}{3} + 1 = \frac{402}{3} + 1 = 134 + 1 = 135\)
Step 2: Find the general terms of both sequences.
AP₁: \(t_m = 3 + (m-1)4 = 4m - 1\), where \(1 \le m \le 101\)
AP₂: \(t_n = 2 + (n-1)3 = 3n - 1\), where \(1 \le n \le 135\)
Step 3: Find the condition for common terms.
Common terms satisfy \(4m - 1 = 3n - 1 \Rightarrow 4m = 3n \Rightarrow m = \frac{3n}{4}\)
Since \(m\) must be an integer, \(n\) must be a multiple of 4.
Let \(n = 4k\), then \(m = 3k\).
Step 4: Determine the range of \(k\).
From AP₁: \(1 \le m \le 101 \Rightarrow 1 \le 3k \le 101 \Rightarrow 1 \le k \le \lfloor \frac{101}{3} \rfloor = 33\)
From AP₂: \(1 \le n \le 135 \Rightarrow 1 \le 4k \le 135 \Rightarrow 1 \le k \le \lfloor \frac{135}{4} \rfloor = 33\)
So \(k = 1, 2, 3, \dots, 33\)
Step 5: Find the common terms.
Common term for a given \(k\) is: \(t = 3n - 1 = 3(4k) - 1 = 12k - 1\)
So the common terms form an AP: 11, 23, 35, ..., with first term \(A = 11\), common difference \(D = 12\), and 33 terms.
Step 6: Find the last term and verify.
Last term when \(k = 33\): \(12 \times 33 - 1 = 396 - 1 = 395\)
Check in AP₁: \(395 = 4m - 1 \Rightarrow m = 99\) (valid since \(1 \le m \le 101\))
Check in AP₂: \(395 = 3n - 1 \Rightarrow n = 132\) (valid since \(1 \le n \le 135\))
Step 7: Calculate the sum of the common terms.
Sum of an AP = \(\frac{\text{number of terms}}{2} \times (\text{first term} + \text{last term})\)
\[ S = \frac{33}{2} \times (11 + 395) = \frac{33}{2} \times 406 = 33 \times 203 = 6699 \]
Hence, the sum of the common terms is 6699.
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