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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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chetan2u
Gnpth
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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let A1 = a
A2 = a + c
A3 = a + 2c ( each term after the first term is equal to the preceding term plus a constant c.)
.
.
An = a + (n-1)xc ( Nth Term )

A1 = a ,
A11 = a +(11-1)xc = a +10c ,
A21 = a +(21-1)xc = a +20c
A1 +A11+A21 = 3a + 30c = 99
a + 10c = 33
A3 + A19 = a + 2c + a + 18c
A3 + A19 = 2a +20c
A3 + A19 = 2(a +10c)
A3 + A19 = 2 x 33
A3 + A19 = 66
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chetan2u Could you please clarify why you assume that c is 1? I tried the same methodology but got stuck because my equation was 3a + 30c = 99 so I had two variables.

My mentality was
a2 = a1 + c
a3 = a1 + c + c (so a1 +2c)
a4 = a1 + c + c + c (so a1 + 3c)
etc
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a1 = x
a2 = x + c
a3 = x +2c
and so on
so a1 + a11 + a21 = x + x +10c + x + 20c = 99
3x + 30c = 99
x + 10c = 33

a3 + a19 = x + 2c + x +18c = 2x +20c
if x+10c = 33, 2x + 20c will be twice that, so 66

Answer A
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