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Find the last common term in both of the following arithmetic sequences: Sequence 1: 1, 11, 21, 31,... (100 terms) Sequence 2: 31, 36, 41, 46,... (100 terms)

1. 381
2. 521
3. 281
4. None of these

Correct answer: 521

Solution

Sequence 1: first term = 1, common difference = 10. nth term = 1 + 10(n-1). For n = 100: last term = 1 + 990 = 991. Sequence 2: first term = 31, common difference = 5. mth term = 31 + 5(m-1). For m = 100: last term = 31 + 495 = 526. Common terms must satisfy: 1 + 10a = 31 + 5b => 10a - 5b = 30 => 2a - b = 6. The common terms form an AP with first term = 31 (since 31 is in both: 31 = 1+10*3 and 31 = 31+5*0) and common difference = lcm(10,5) = 10. Common AP: 31, 41, 51,..., up to min(991, 526) = 526. Terms: 31 + 10k <= 526 => k <= 49.5, so k = 0,1,...,49, giving 50 terms. Last common term = 31 + 10*49 = 31 + 490 = 521. But 521 <= 526 (yes, in Seq 2) and 521 = 1 + 10*52 = 521 (yes, in Seq 1 since 52 <= 99). So last common term = 521. Wait - let me verify 521 is in Seq 1: 521 = 1 + 10k => k = 52 (term 53, within 100 terms). Yes! And 521 in Seq 2: 521 = 31 + 5k => k = 98 (term 99, within 100 terms). Yes! So last common term = 521.

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