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Number of common terms in both sequence 3, 7, 11, ………….407 and 2, 9, 16, ……..905 is |
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Answer» First common term = 23 common difference = 7 × 4 = 28 Last term ≤ 407 ⇒ 23 + (n –1) × 28 ≤ 407 ⇒ (n –1) × 28 ≤ 384 ⇒ n ≤ 13.71 + 1 n ≤ 14.71 So n = 14 |
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