1.

Number of common terms in both sequence 3, 7, 11, ………….407 and 2, 9, 16, ……..905 is

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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