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# In a sequence of numbers t1, t2, t3, ... the difference of any two suc

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Joined: 02 Sep 2009
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In a sequence of numbers t1, t2, t3, ... the difference of any two suc  [#permalink]

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15 May 2020, 08:16
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78% (01:54) correct 22% (02:14) wrong based on 18 sessions

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In a sequence of numbers $$t_1$$, $$t_2$$, $$t_3$$, ... the difference of any two successive terms is a constant. If $$|t_8| = |t_{16}|$$ and $$t_3 ≠ t_7$$, what is the sum of the first 23 terms of the sequence?

A. -10
B. -4
C. 0
D. 4
E. 10

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Re: In a sequence of numbers t1, t2, t3, ... the difference of any two suc  [#permalink]

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15 May 2020, 10:07
1
nth term of the sequence = a+(n-1)d

a= first term
d= the difference of any two successive terms

$$t_3 ≠ t_7$$

$$a+2d ≠ a+6d$$

d≠0

$$|t_8| = |t_{16}|$$

$$|a+7d| = |a+15d|$$

$$Case 1-$$ a+7d= a+15d

We get d=0 (not possible)

Reject this case

Case 2- a+7d= -(a+15d)

2a+22d=0 or a+11d =0

Sum of first 23 terms = Mean of 23 terms * 23

Mean of 23 terms = 12th term of the sequence = a+11d

Sum of first 23 terms = (a+11d) * 23 = 0

Bunuel wrote:
In a sequence of numbers $$t_1$$, $$t_2$$, $$t_3$$, ... the difference of any two successive terms is a constant. If $$|t_8| = |t_{16}|$$ and $$t_3 ≠ t_7$$, what is the sum of the first 23 terms of the sequence?

A. -10
B. -4
C. 0
D. 4
E. 10

Are You Up For the Challenge: 700 Level Questions
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Re: In a sequence of numbers t1, t2, t3, ... the difference of any two suc  [#permalink]

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15 May 2020, 17:07
In a sequence of numbers t1t1, t2t2, t3t3, ... the difference of any two successive terms is a constant. If |t8|=|t16| and t3≠t7, what is the sum of the first 23 terms of the sequence?

Tn = a + (n - 1) d, where Tn = nth term and a = first term. Here d = common difference = Tn - Tn-1.
Sn =(n/2)[2a + (n- 1)d]

given,|t8|=|t16I
or,|a+7d|=|a+15d|
case I : a+7d= a+15d
or, d=0 which is not possible because if common difference is zero each term is equal which invalidates t3≠t7 .

so, case II : a+7d= -(a+15d)
or, 2a + 22d = 0

now, the sum of the first 23 terms of the sequence
=(23/2)[2a + (23- 1)d]
=23/2[2a + 22d ]
=0

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Re: In a sequence of numbers t1, t2, t3, ... the difference of any two suc  [#permalink]

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16 May 2020, 08:17
Bunuel wrote:
In a sequence of numbers $$t_1$$, $$t_2$$, $$t_3$$, ... the difference of any two successive terms is a constant. If $$|t_8| = |t_{16}|$$ and $$t_3 ≠ t_7$$, what is the sum of the first 23 terms of the sequence?

A. -10
B. -4
C. 0
D. 4
E. 10

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The scenario here is either $$t_8 = -t_{16}$$, or all terms have a difference of 0. Since $$t_3 ≠ t_7$$ we clearly cannot have a difference of 0, thus we can conclude $$t_8 = -t_{16}$$.

Note there are no actual numbers present in this question, an educated guess would be to choose C since we don't have any numbers to conclude an answer like 4 or 10.

In fact we do have $$t_9 = -t_{15}$$, $$t_{10} = -t_{14}$$, and $$t_{11} = -t_{13}$$, so we must have $$t_{12} = 0$$. Any sum centered around $$t_{12}$$ would also be equivalent to zero, the terms 1 to 23 have a center of 12 so that sum is indeed equal to 0.

Ans: C
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In a sequence of numbers t1, t2, t3, ... the difference of any two suc  [#permalink]

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16 May 2020, 13:17
Bunuel wrote:
In a sequence of numbers $$t_1$$, $$t_2$$, $$t_3$$, ... the difference of any two successive terms is a constant. If $$|t_8| = |t_{16}|$$ and $$t_3 ≠ t_7$$, what is the sum of the first 23 terms of the sequence?

A. -10
B. -4
C. 0
D. 4
E. 10

Are You Up For the Challenge: 700 Level Questions

If |t8|=|t16|, then logically one of them is negative and other one is positive, with equal value
In such case, the middle value between t8 and t16, which is t12, must be 0 (this middle value basically indicates the transition from negative to positive or vice versa)

Now, S23 = (23/2) x (2a + 22d) = 23(a + 11d) = 23 x t12 = 0

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In a sequence of numbers t1, t2, t3, ... the difference of any two suc   [#permalink] 16 May 2020, 13:17

# In a sequence of numbers t1, t2, t3, ... the difference of any two suc

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