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# term after the first term is equal to the preceding term

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CEO
Joined: 07 Jun 2014
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term after the first term is equal to the preceding term [#permalink]  18 Jan 2016, 02:25
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a1, a2, a3 ................ an

In the sequence above, each term after the first term is equal to the preceding term plus the constant c. If a1+a3+a5=27, what is the value of a2+a4 ?

a2+a4=

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Sandy
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CEO
Joined: 07 Jun 2014
Posts: 2549
GRE 1: 323 Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 41

Kudos [?]: 734 [2] , given: 127

Re: term after the first term is equal to the preceding term [#permalink]  18 Jan 2016, 04:52
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Solution

This is clearly an arithmetic progression and we just need the first 5 terms. We can rewrite the progression as a - 2c, a - c, a , a + c, a + 2c. Here the difference between the terms is c as described in the problem

a1 + a3 + a5 = a - 2c + a + a+ 2c = 3a = 27. Hence a =9.

To find a2 + a4 = a - c + a + C = 2a =18.

The purpose of writing this series in the above form is to eliminate c as a variable. Hence answer is 18.
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Re: term after the first term is equal to the preceding term [#permalink]  30 Mar 2016, 09:05
i don't understand where you derive a - 2c and a + 2c from
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Re: term after the first term is equal to the preceding term [#permalink]  01 Apr 2016, 02:30
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Lets take a real arithmetic progression:

1, 5, 9 , 13, 17

As you can see any term of the above arithmetic progression is 4 greater than the previous term. 1+4 = 5... 5+4 = 9.... so on

So if we draw parallels with the question above here c = 4. Then we can write the above series taking the middle term 9 into account:
9-2*4, 9-4, 9 , 9+4, 9+2*4
a-2*c, a-c , a , a+c, a+2*c

Thats how Sandy derives the a+2c and a-2c terms
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Re: term after the first term is equal to the preceding term [#permalink]  29 Jun 2017, 20:10
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If that is confusing then, you can say a, a+c, a+2c, a+3c, a+4c

a1+a3+a5 = 27

a + a+2c + a+4c = 27
a = 9-2c

now a2 + a4

a+c + a+3c
2a+4c
2(9-2c) + 4c
18-4c+4c

Re: term after the first term is equal to the preceding term   [#permalink] 29 Jun 2017, 20:10
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