Question:
Let \(a_n=46+8n\) and \(b_n=98+4n\) be two sequences for natural numbers \(n ≤ 100\) . Then, the sum of all terms common to both the sequences is
Let \(a_n=46+8n\) and \(b_n=98+4n\) be two sequences for natural numbers \(n ≤ 100\) . Then, the sum of all terms common to both the sequences is
Updated On: Aug 17, 2024
- 14900
- 15000
- 14798
- 14602
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The Correct Option is A
Solution and Explanation
The first series goes as follows: \( 46, 54, 62, 70, 78, 86, 94, 102,.... \)
The second series follows this format: Numbers \(98, 102, 106, 110,..\)
The difference between them will be hcf(4,8) = 8, which is the first common term (term of the common terms).
The necessary order is \(102,110,118\), (the final term must be less than 468, which is the second series' hundredth term).
\([102 + (n-1)(8) ≤ 498]\)
If n is equal to or less than 50.5, then n = 50.
Applying the A.P. formula summation, we get: Required sum = \(\frac{n}{2}(2a + (n - 1)d) = \frac{50}{2}(2 \times 102 + 49 \times 8) = 14900\)
The correct option is (A): 14900.
The second series follows this format: Numbers \(98, 102, 106, 110,..\)
The difference between them will be hcf(4,8) = 8, which is the first common term (term of the common terms).
The necessary order is \(102,110,118\), (the final term must be less than 468, which is the second series' hundredth term).
\([102 + (n-1)(8) ≤ 498]\)
If n is equal to or less than 50.5, then n = 50.
Applying the A.P. formula summation, we get: Required sum = \(\frac{n}{2}(2a + (n - 1)d) = \frac{50}{2}(2 \times 102 + 49 \times 8) = 14900\)
The correct option is (A): 14900.
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