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In a sequence of numbers in which each term after the first is found b

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In a sequence of numbers in which each term after the first is found b  [#permalink]

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New post 15 May 2020, 08:37
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In a sequence of numbers in which each term after the first is found by multiplying the previous one by a non-zero constant, the sum of the first 12 terms is equal to the sum of the first 14 terms. If the sum of the first 17 terms is 92, what is the third term of the sequence?

A. -92
B. -46
C. 46
D. 92
E. 231



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In a sequence of numbers in which each term after the first is found b  [#permalink]

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New post 15 May 2020, 09:45
Sum of first 12 terms \(= \frac{a(r^{12}-1)}{(r-1)}\)

Sum of first 14 terms \(= \frac{a(r^{14}-1)}{(r-1)}\)

\(\frac{a(r^{12}-1)}{(r-1)} = \frac{a(r^{14}-1)}{(r-1)}\)

\(r^{14}=r^{12}\)

r= 0(not possible), 1(not possible) or -1

r=-1

\(\frac{a(r^{17}-1)}{(r-1)} = 92\)

\(a*\frac{-2}{-2} = 92\)

a=92

\(ar^2= 92(-1)^2 = 92\)

Alternate solution-
The sum of the first 12 terms is equal to the sum of the first 14 terms is only possible when 2 subsequent terms are equal in magnitude and opposite in sign.

Each odd number term is same.

the sum of the first 17 terms is 92; 17th term must be equal to 92 and so does the third term.



Bunuel wrote:
In a sequence of numbers in which each term after the first is found by multiplying the previous one by a non-zero constant, the sum of the first 12 terms is equal to the sum of the first 14 terms. If the sum of the first 17 terms is 92, what is the third term of the sequence?

A. -92
B. -46
C. 46
D. 92
E. 231



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Re: In a sequence of numbers in which each term after the first is found b  [#permalink]

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New post 15 May 2020, 17:20
In a sequence of numbers in which each term after the first is found by multiplying the previous one by a non-zero constant, the sum of the first 12 terms is equal to the sum of the first 14 terms. If the sum of the first 17 terms is 92, what is the third term of the sequence?

Sum of G.P. up to n terms if 1st term is a and common ratio is r .
Sn =a.(1-r^n)/(1-r)………….(1) while r < 1 but (r ≠ 1)
and,Sn =a.(r^n -1)/(r - 1)………(2) while r > 1 but (r ≠ 1)

given, S12=S14
or,a.(1-r^12)/(1-r)=a.(1-r^14)/(1-r) (assuming, r<1)
or,r^12=r^14
so, r=-1

now,the sum of the first 17 terms is 92
a.(1-r^17)/(1-r) = 92
or, a= 92*2/(1+1)
a=92

the third term of the sequence= ar^2 = 92

correct answer D
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Re: In a sequence of numbers in which each term after the first is found b  [#permalink]

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New post 15 May 2020, 21:53
In a sequence of numbers in which each term after the first is found by multiplying the previous one by a non-zero constant
\( S = x, c*x, c^2*x, ........... \)

the sum of the first 12 terms is equal to the sum of the first 14 terms

\( S_{12} = S_{14} \)
\( x+cx+c^2x+.......c^{11}x = x+cx+c^2x+.......c^{11}x + c^{12}x + c^{13}x \)
\( c^{12}x + c^{13}x = 0 \)
\( c^{12}x (1+c) = 0 \)
\( c = -1, or, c^{12}x = 0 \)
\( from, c^{12}x = 0, if x = 0, then \)
sum of the first 17 terms is not equal to 92,
so, x is not 0 and c = -1.
Now, considering c = -1, sum of first 17 terms = 92
\( x -x + x - x +...........-x (16^{th} term) + x (17^{th} term) = 92 \)
only 17th term remains, giving x = 92
Now, the third term of the above sequence is x, which is equal to 92.
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Re: In a sequence of numbers in which each term after the first is found b   [#permalink] 15 May 2020, 21:53

In a sequence of numbers in which each term after the first is found b

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