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In the sequence a1, a2, ... , an each term after the first term is
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Gnpth
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In the sequence a1, a2, ... , an each term after the first term is [#permalink] New post  Updated on: 11 Nov 2017, 17:44
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In the sequence \(a_1\), \(a_2\), ... ,\(a_n\) each term after the first term is equal to the preceding term plus a constant c.

\(a_1\) + \(a_{11}\) + \(a_{21}\) = 99. What is the value of \(a_3\)+\(a_{19}\)?

A. 66
B. 44
C. 33
D. 22
E. 11
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Originally posted by Gnpth on 10 Nov 2017, 18:26.
Last edited by Gnpth on 11 Nov 2017, 17:44, edited 2 times in total.
formatted the choices
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chetan2u
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Re: In the sequence a1, a2, ... , an each term after the first term is [#permalink] New post  10 Nov 2017, 19:09
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Gnpth wrote:
In the sequence \(a_1\), \(a_2\), ... ,\(a_n\) each term after the first term is equal to the preceding term plus a constant c.

\(a_1\) + \(a_{11}\) + \(a_{21}\) = 99. What is the value of \(a_3\)+\(a_{19}\)?

A. 33
B. 66
C. 22
D. 44
E. 11


Hi..

As each term is C more than the previous one...
Each term, \(a_x\) will be (x-1) times c more than \(a_1\)
So \(a_1+a_{11}+a_{21}=a_1+a_1+(11-1)+a_1+(21-1)=3*a_1+30=99....a_1=23\)

Now \(a_3+a_{19}=a_1+(3-1)+a_1+(19-1)=2*a_1+20=2*23+20=66\)

B

But the choices require formatting as per the GMAT. All choices are to be some order then 66 becomes choice A
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Re: In the sequence a1, a2, ... , an each term after the first term is [#permalink] New post  10 Nov 2017, 19:26
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Gnpth wrote:
In the sequence \(a_1\), \(a_2\), ... ,\(a_n\) each term after the first term is equal to the preceding term plus a constant c.

\(a_1\) + \(a_{11}\) + \(a_{21}\) = 99. What is the value of \(a_3\)+\(a_{19}\)?

A. 66
B. 44
C. 33
D. 22
E. 11


a2 = a1 + c, a3 = a2+c = a1+2c
a11 = a1+10C, a21 = a1+20c
3a1 + 30c = 99, a1+10c =33

a3+a19 = 2a1+20c = 2(a1+10c)=66
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jfranciscocuencag
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In the sequence a1, a2, ... , an each term after the first term is [#permalink] New post  18 Feb 2019, 17:39
1
Pn = Pn-1 + C (constant)

It is an arithmetic progression:

Sum = (Average)(n)

A1 + A11 + A21 = 99

\(\frac{99}{3} = 33\) (Average)

Sum = (33)(2) ... 2 terms (A3 + A19)

Sum = 66

A
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Re: In the sequence a1, a2, ... , an each term after the first term is [#permalink] New post  04 Jul 2019, 04:40
chetan2u wrote:
Gnpth wrote:
In the sequence \(a_1\), \(a_2\), ... ,\(a_n\) each term after the first term is equal to the preceding term plus a constant c.

\(a_1\) + \(a_{11}\) + \(a_{21}\) = 99. What is the value of \(a_3\)+\(a_{19}\)?

A. 33
B. 66
C. 22
D. 44
E. 11


Hi..

As each term is C more than the previous one...
Each term, \(a_x\) will be (x-1) times c more than \(a_1\)
So \(a_1+a_{11}+a_{21}=a_1+a_1+(11-1)+a_1+(21-1)=3*a_1+30=99....a_1=23\)

Now \(a_3+a_{19}=a_1+(3-1)+a_1+(19-1)=2*a_1+20=2*23+20=66\)

B

But the choices require formatting as per the GMAT. All choices are to be some order then 66 becomes choice A



Why C has been assumed as 1?

Thanks
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