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Re: In the sequence above, each term is 9 more than the previous term. Wha [#permalink]
First term , a= 14
Common difference , d = 9
nth term , tn = a + (n-1)d
41st term , t41 = a+ 40*d = 14 + 40*9 = 374

Answer D
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Re: In the sequence above, each term is 9 more than the previous term. Wha [#permalink]
The numbers in this sequence end with 4,3,2,1,0,9,8,7,6,5,4,3,2,1......
In other words, after 10 terms, the units digits will repeat. So the 41st term will be a value ending in 4. Only answer is D
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Re: In the sequence above, each term is 9 more than the previous term. Wha [#permalink]
\(a = 14, d = 9\)
41st term = \(a + 40d = 14 + 40(9) = 14 + 360 = 374\). Ans - D.
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Re: In the sequence above, each term is 9 more than the previous term. Wha [#permalink]
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Re: In the sequence above, each term is 9 more than the previous term. Wha [#permalink]
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