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# In the sequence above, each term is 9 more than the previous term. Wha

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Posts: 50002
In the sequence above, each term is 9 more than the previous term. Wha  [#permalink]

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30 Nov 2015, 02:36
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14, 23, 32, 41, 50, 59, …

In the sequence above, each term is 9 more than the previous term. What is the 41st term of the sequence?

A) 360

B) 365

C) 369

D) 374

E) 383

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Re: In the sequence above, each term is 9 more than the previous term. Wha  [#permalink]

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30 Nov 2015, 03:05
41st term = a + 40d
a = 14; d = 9
41st term = 14 + 40*9 = 374

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Re: In the sequence above, each term is 9 more than the previous term. Wha  [#permalink]

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30 Nov 2015, 04:20
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

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Re: In the sequence above, each term is 9 more than the previous term. Wha  [#permalink]

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30 Nov 2015, 11:13
Bunuel wrote:
14, 23, 32, 41, 50, 59, …

In the sequence above, each term is 9 more than the previous term. What is the 41st term of the sequence?

A) 360

B) 365

C) 369

D) 374

E) 383

The above posts apply a formula for arithmetic sequences.
However, if you don't know that formula, you can also solve the question by looking for a pattern.

term1 = 14
term2 = 14 + 9
term3 = 14 + 9 + 9
term4 = 14 + 9 + 9 + 9
term5 = 14 + 9 + 9 + 9 + 9
.
.
.
.
term41 = 14 + (sum of forty 9's)
= 14 + (360)
= 374
= D

Cheers,
Brent
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Re: In the sequence above, each term is 9 more than the previous term. Wha  [#permalink]

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29 Jun 2017, 19:36
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]

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29 Jun 2017, 22:03
$$a = 14, d = 9$$
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 &nbs [#permalink] 29 Jun 2017, 22:03
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