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In a sequence of four positive integers, the first three terms are in

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Bunuel
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In a sequence of four positive integers, the first three terms are in [#permalink] In a sequence of four positive integers, the first three terms are in   14 May 2020, 03:35
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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


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Re: In a sequence of four positive integers, the first three terms are in [#permalink] In a sequence of four positive integers, the first three terms are in   14 May 2020, 06:04
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Bunuel
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


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
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nick1816
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In a sequence of four positive integers, the first three terms are in [#permalink] In a sequence of four positive integers, the first three terms are in   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


Bunuel
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
­
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Re: In a sequence of four positive integers, the first three terms are in [#permalink] In a sequence of four positive integers, the first three terms are in   14 May 2020, 20:41
chetan2u
Bunuel
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


Let the terms be a, a+d, a+2d, b, but b=a+40
So, the sequence = a, a+d, a+2d, a+40

Last 3 terms are in GP...
So, \(\frac{a+40}{a+2d}=\frac{a+2d}{a+d}.........(a+40)(a+d)=(a+2d)(a+2d)\).....
\(a^2+ad+40a+40d=a^2+4ad+4d^2.......40a+40d=3ad+4d^2\)......
\(40a-3ad=4d^2-40d.....a(40-3d)=4d(d-10)\)....
\(a=\frac{4d(d-10)}{40-3d}\)

Now, a>0, so \(\frac{4d(d-10)}{40-3d}>0\)
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
II) At d=10, a=0..Not possible
Thus d>10
III) But denominator becomes negative when \(40<3d\) or \(d>13.33\)

So, \(10<d<13.33\) or \(d=11, 12, 13\)

Now ONLY d= 12 gives a as a positive integer, so d=12 and \(a=\frac{4d(d-10)}{40-3d}=\frac{4*12(12-10)}{40-3*12}=\frac{4*24}{4}=24\)

Sequence {a, a+d, a+2d, a+40} = { 24, 36, 48, 64}

SUM = 24+36+48+64=172

D

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!
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In a sequence of four positive integers, the first three terms are in [#permalink] In a sequence of four positive integers, the first three terms are in   14 May 2020, 20:43
nick1816
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 or 12

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
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
Hi nick1816,

If \(10<d<\frac{40}{3}\), then this means that d = 11, 12 and 13. 13 is also a valid value of d.­
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Re: In a sequence of four positive integers, the first three terms are in [#permalink] In a sequence of four positive integers, the first three terms are in   14 May 2020, 21:36
Expert Reply
abcdddddd
chetan2u
Bunuel
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


Let the terms be a, a+d, a+2d, b, but b=a+40
So, the sequence = a, a+d, a+2d, a+40

Last 3 terms are in GP...
So, \(\frac{a+40}{a+2d}=\frac{a+2d}{a+d}.........(a+40)(a+d)=(a+2d)(a+2d)\).....
\(a^2+ad+40a+40d=a^2+4ad+4d^2.......40a+40d=3ad+4d^2\)......
\(40a-3ad=4d^2-40d.....a(40-3d)=4d(d-10)\)....
\(a=\frac{4d(d-10)}{40-3d}\)

Now, a>0, so \(\frac{4d(d-10)}{40-3d}>0\)
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
II) At d=10, a=0..Not possible
Thus d>10
III) But denominator becomes negative when \(40<3d\) or \(d>13.33\)

So, \(10<d<13.33\) or \(d=11, 12, 13\)

Now ONLY d= 12 gives a as a positive integer, so d=12 and \(a=\frac{4d(d-10)}{40-3d}=\frac{4*12(12-10)}{40-3*12}=\frac{4*24}{4}=24\)

Sequence {a, a+d, a+2d, a+40} = { 24, 36, 48, 64}

SUM = 24+36+48+64=172

D

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!


Yes, you are correct and that’s what I meant by ‘only 12 as the answer’. But missed out writing that ‘fits in’. Now adding that. Thanks.
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In a sequence of four positive integers, the first three terms are in [#permalink] In a sequence of four positive integers, the first three terms are in   14 May 2020, 21:43
You could infer that from the last line. Anyways Edited! Thanks.

abcdddddd
nick1816
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 or 12

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
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
Hi nick1816,

If \(10<d<\frac{40}{3}\), then this means that d = 11, 12 and 13. 13 is also a valid value of d.
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Re: In a sequence of four positive integers, the first three terms are in [#permalink] In a sequence of four positive integers, the first three terms are in   21 Sep 2024, 03:21
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Re: In a sequence of four positive integers, the first three terms are in [#permalink]
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