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# In a certain series, each term is m greater than the previous term. If

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In a certain series, each term is m greater than the previous term. If [#permalink]

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22 Jun 2015, 11:44
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In a certain series, each term is m greater than the previous term. If the 17th term is 560 and the 14th term is 500, what is the first term?

A. 220
B. 240
C. 260
D. 290
E. 305

[Reveal] Spoiler:
In this arithmetic set, the difference between any two consecutive terms is m. Since there are three terms from a14 to a17 we can conclude that:

--> a17 = a14+3·m
--> 560 = 500+3·m
--> m=20

We now can conclude that:

--> a1 = a14-13·m
--> a1 = 500-13·20
--> a1 = 240
[Reveal] Spoiler: OA

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Re: In a certain series, each term is m greater than the previous term. If [#permalink]

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22 Jun 2015, 19:43
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reto wrote:
In a certain series, each term is m greater than the previous term. If the 17th term is 560 and the 14th term is 500, what is the first term?

A. 220
B. 240
C. 260
D. 290
E. 305

[Reveal] Spoiler:
In this arithmetic set, the difference between any two consecutive terms is m. Since there are three terms from a14 to a17 we can conclude that:

--> a17 = a14+3·m
--> 560 = 500+3·m
--> m=20

We now can conclude that:

--> a1 = a14-13·m
--> a1 = 500-13·20
--> a1 = 240

Another method is make equations and solve simultaneously:

a + 16m = 560 (17th term)
a + 13m = 500 (14th term)

You get a = 240 and m = 20
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In a certain series, each term is m greater than the previous term. If [#permalink]

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22 Jun 2015, 21:47
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reto wrote:
In a certain series, each term is m greater than the previous term. If the 17th term is 560 and the 14th term is 500, what is the first term?

A. 220
B. 240
C. 260
D. 290
E. 305

The detailed explanation

CONCEPT: The series where each terms is more/less than previous term by a constant amount (common difference) resulting in an increasing or decreasing series is called an Arithmetic Progression.

i.e. If First term = a and the common difference = d

then the terms in the series can be written as

a, (a+d), (a+2d), (a+3d), (a+4d), (a+5d), (a+6d), (a+7d)...

First term, $$T_1 = a$$
Second term, $$T_2 = (a+d)$$
Third term, $$T_3 = (a+2d)$$
Forth term, $$T_4 = (a+3d)$$
...
i.e. nth Term of the series,$$T_n = a+(n-1)d$$

Here the question has mentioned that common difference = m

i.e. 17th Term, $$T_{17} = a+(17-1)m = a+16m = 560$$ -------- Equation (1)
and 14th Term, $$T_{14} = a+(14-1)m = a+13m = 500$$ -------- Equation (2)

Subtract the second equation from the first
i.e. $$(a+16m) - (a+13m) = 560 - 500$$

i.e. $$3m = 60$$
i.e. $$m = 20$$

Substituting in Equation (2), we get
$$a + 13*20 = 500$$
i.e. $$a = 500 - 260 = 240$$

Answer: option
[Reveal] Spoiler:
B

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Re: In a certain series, each term is m greater than the previous term. If [#permalink]

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07 Oct 2017, 02:22
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Re: In a certain series, each term is m greater than the previous term. If   [#permalink] 07 Oct 2017, 02:22
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# In a certain series, each term is m greater than the previous term. If

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