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14 May 2020, 03:35
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Difficulty:
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In a sequence of four positive integers, the first three terms are in A.P and the last three terms are in G.P. If the difference between the first and last terms is 40, what is the sum of all four terms of the sequence?
A. 108
B. 124
C. 136
D. 172
E. 196
Are You Up For the Challenge: 700 Level Questions
A. 108
B. 124
C. 136
D. 172
E. 196
Are You Up For the Challenge: 700 Level Questions
14 May 2020, 06:04
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Bunuel
Let the terms be \(a, a+d, a+2d, b\), but b=a+40
So, the sequence = \(a, a+d, a+2d, a+40\)
We are given that last 3 terms are in GP...
So, a+d, a+2d, a+40 are in GP, and that will give us \(\frac{a+40}{a+2d}=\frac{a+2d}{a+d}\).....
Let us simplify it now
\((a+40)(a+d)=(a+2d)(a+2d)\).....
\(a^2+ad+40a+40d=a^2+4ad+4d^2\).....
\(40a+40d=3ad+4d^2\)......
\(40a-3ad=4d^2-40d\).....
\(a(40-3d)=4d(d-10)\)....
\(a=\frac{4d(d-10)}{40-3d}\)
Now, we are given that sequence has positive integers, so a>0, and this further tells us that \(\frac{4d(d-10)}{40-3d}>0\).
Value of a is dependent on value of d, so let us check d for the values that keep a as positive
Case I :- when d<10, Numerator=\(4d(d-10)<0\), but denominator = \(40-3d>0\), so \(\frac{N}{D}=\frac{(-)}{(+)}=(-)\), but a>0...so NOT possible
Case II :- At d=10, a=0..Not possible
Thus, from I and II, we get \(d>10\)
Case III :- But denominator becomes negative when \(40<3d\) or \(d>13.33\)
So, The combined cases above give us \(10<d<13.33\) or \(d=11, 12, 13\)
Now ONLY d= 12 gives a as a positive integer and fits in the answer choices, so d=12 and \(a=\frac{4d(d-10)}{40-3d}=\frac{4*12(12-10)}{40-3*12}=\frac{4*24}{4}=24\)
# Although 13 gives you an integer value for a but does not match with the choices, when you substitute d as 13.
Sequence {a, a+d, a+2d, a+40} = { 24, 36, 48, 64}
SUM = \(24+36+48+64=172\)
D
General Discussion
14 May 2020, 06:35
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sequence is
a, a+d, a+2d, a+40
Now since last 3 terms are in GP,
\((a+2d)^2 = (a+d)(a+40)\)
\(a^2+4d^2+4ad = a^2+ad+40a+40d\)
\(4d^2-40d=40a-3ad\)
\(4d(d-10) = a(40-3d)\)
\(a=\frac{4d(d-10)}{(40-3d)}\)
since a>0, either \(d<0\) or \(10<d<\frac{40}{3}\)
Case 1- d<0
\(c= \frac{4d(d-10)}{40-3d)} +2d\)
\(c= \frac{4d^2-40d+80d-6d^2}{(40-3d)}\)
\(c= \frac{-2d^2+40d}{(40-3d)} <0\)
(reject this case)
Case 2-
\(10<d<\frac{40}{3}\)
Since each term is integer, d must be an integer.
d can be 11,12 or 13
1. d= 11; a is not an integer (Reject it)
2. d=12; a=24
b= 24+12=36; c= 36+12=48; d=24+40=64
Also, \(\frac{48}{36} = \frac{64}{48} = \frac{4}{3}\)
a+b+c+d= 24+36+48+64 = 172
don't need to check further
a, a+d, a+2d, a+40
Now since last 3 terms are in GP,
\((a+2d)^2 = (a+d)(a+40)\)
\(a^2+4d^2+4ad = a^2+ad+40a+40d\)
\(4d^2-40d=40a-3ad\)
\(4d(d-10) = a(40-3d)\)
\(a=\frac{4d(d-10)}{(40-3d)}\)
since a>0, either \(d<0\) or \(10<d<\frac{40}{3}\)
Case 1- d<0
\(c= \frac{4d(d-10)}{40-3d)} +2d\)
\(c= \frac{4d^2-40d+80d-6d^2}{(40-3d)}\)
\(c= \frac{-2d^2+40d}{(40-3d)} <0\)
(reject this case)
Case 2-
\(10<d<\frac{40}{3}\)
Since each term is integer, d must be an integer.
d can be 11,12 or 13
1. d= 11; a is not an integer (Reject it)
2. d=12; a=24
b= 24+12=36; c= 36+12=48; d=24+40=64
Also, \(\frac{48}{36} = \frac{64}{48} = \frac{4}{3}\)
a+b+c+d= 24+36+48+64 = 172
don't need to check further
Bunuel
14 May 2020, 20:41
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chetan2u
Hi chetan2u,
According to your explanation, d = 11,12,13
For d = 13, a is also a positive integer because denominator of a becomes 40-3(d) = 40 - 3 (13) = 40 - 39 = 1
So, d = 13 is also a possibility. Of course, if we consider d = 13 then the sum of the numbers will be greater than the answer choices. But you mentioned that d = 12 is the ONLY value for which a is an integer. That's not true!
