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In a sequence of numbers in which the difference of any two successive

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Joined: 02 Sep 2009
Posts: 65807
In a sequence of numbers in which the difference of any two successive  [#permalink]

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15 May 2020, 07:40
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45% (medium)

Question Stats:

71% (02:20) correct 29% (02:22) wrong based on 24 sessions

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In a sequence of numbers in which the difference of any two successive terms is a constant, 4 times the 4th term is equal to 9 times the 9th term, what is 13 times the 13th term ?

A. 0
B. 5
C. 12
D. 13
E. 15

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Re: In a sequence of numbers in which the difference of any two successive  [#permalink]

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15 May 2020, 12:23
1
Bunuel wrote:
In a sequence of numbers in which the difference of any two successive terms is a constant, 4 times the 4th term is equal to 9 times the 9th term, what is 13 times the 13th term ?

A. 0
B. 5
C. 12
D. 13
E. 15

an= a+(n-1)*d
a4 = a1+(3)*d and a9 = a1+8d
given
4*a4 = 9*a9
so we can write as
4a1+12d = 9a1+72d
solve for a1 ; -5a1=60d ;
a1 = -12d
so
a13 = a1+12d
a13 = -12d+12d ; 0
OPTION A
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Re: In a sequence of numbers in which the difference of any two successive  [#permalink]

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15 May 2020, 07:47
Tn = a + (n - 1) d, where Tn = nth term and a = first term. Here d = common difference = Tn - Tn-1.
now,4T4=9T9
or, 4[a+3d]=9[a+8d]
or,-5a=60d
or,a=-12d
or,a+12d=0

now,13 times the 13th term
=13T13
=13[a+12d]
=13*0
=0

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Re: In a sequence of numbers in which the difference of any two successive  [#permalink]

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16 May 2020, 07:16
given $$4t_4=9t_9$$
rewriting the terms in terms of sequence equation we get
$$4(a+3d)=9(a+8d)$$
solving it we will get $$a+12d=0$$

Now $$a+12d$$= $$t_{13}$$

so if $$t_{13}=0$$ then 13 times $$t_{13}$$ will be zero too.

Ans A
Re: In a sequence of numbers in which the difference of any two successive   [#permalink] 16 May 2020, 07:16