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# In a certain sequence of numbers, a1, a2, a3, ..., an, the

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In a certain sequence of numbers, a1, a2, a3, ..., an, the [#permalink]

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06 Oct 2013, 09:30
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In a certain sequence of numbers, a1, a2, a3, ..., an, the average (arithmetic mean) of the first m consecutive terms starting with a1 is m, for any positive integer m. If a1=1, what is a10?

A. 100
B. 55
C. 21
D. 19
E. 1
[Reveal] Spoiler: OA
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Re: In a certain sequence of numbers, a1,a2,a3,...,an, the avera [#permalink]

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06 Oct 2013, 09:45
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nave81 wrote:
In a certain sequence of numbers, a1,a2,a3,...,an, the average (arithmetic mean) of the first m consecutive terms starting with a1 is m, for any positive integer m. If a1=1, what is a10?

A. 100
B. 55
C. 21
D. 19
E. 1

I got D: 19.

As stated above, (a1+a2+a3......+a10)/10=10
therefore a1+a2+a3.......a10=100 (1)
using the same logic, we got a1+a2+a3..........+a9=81 (2)
(2)-(1) we got a10=19
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Re: In a certain sequence of numbers, a1,a2,a3,...,an, the avera [#permalink]

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06 Oct 2013, 09:49
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nave81 wrote:
In a certain sequence of numbers, a1,a2,a3,...,an, the average (arithmetic mean) of the first m consecutive terms starting with a1 is m, for any positive integer m. If a1=1, what is a10?

A. 100
B. 55
C. 21
D. 19
E. 1

The average of the first 9 consecutive terms starting with a1 is 9 --> $$\frac{a_1+a_2+a_3+...+a_{9}}{9}=9$$ --> $$a_1+a_2+a_3+...+a_{9}=81$$.

The average of the first 10 consecutive terms starting with a1 is 10 --> $$\frac{a_1+a_2+a_3+...+a_{10}}{10}=10$$ --> $$a_1+a_2+a_3+...+a_{10}=100$$.

Subtract the first equation from the second: $$a_{10}=100-81=19$$.

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Re: In a certain sequence of numbers, a1,a2,a3,...,an, the avera [#permalink]

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06 Oct 2013, 21:42
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nave81 wrote:
In a certain sequence of numbers, a1,a2,a3,...,an, the average (arithmetic mean) of the first m consecutive terms starting with a1 is m, for any positive integer m. If a1=1, what is a10?

A. 100
B. 55
C. 21
D. 19
E. 1

The average of 1st 10 numbers is 10

Hence,
The average of 5th and 6th term is 10 and therefore 5th term is 9 and 6th term is 11
Extrapolating this trend we get 10th term = 19 i.e. 9,11,13,15,17,19
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Re: In a certain sequence of numbers, a1,a2,a3,...,an, the avera [#permalink]

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06 Oct 2013, 22:17
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nave81 wrote:
In a certain sequence of numbers, a1,a2,a3,...,an, the average (arithmetic mean) of the first m consecutive terms starting with a1 is m, for any positive integer m. If a1=1, what is a10?

A. 100
B. 55
C. 21
D. 19
E. 1

A slight different approach.
Sum of first m numbers is: $$[(First term$$+$$Last term)$$$$/$$$$2]$$*$$m$$.
Here it will be $$(a1+a10)/2$$*$$m$$.
Also, since the average is given as m, therefore the sum would be $$m^2$$.

On solving, $$a1+a10=2m$$
$$a1+a10=20$$ or $$a10=20-1=19$$
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Re: In a certain sequence of numbers, a1,a2,a3,...,an, the avera [#permalink]

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07 Oct 2013, 23:33
Is it appropriate to do some good guess work out by looking at the options?
For example if a1=1, and if a2, a3 ... a10 are in sequence then there are 3 possiblities for the answer of a10.
1st is that a1.. a10 are the first consecutive 10 nos. in that case the ans should be 10 which is not one of the options.
2nd is that a1... a10 could even consecutive nos. in that case nth term equals to 2n, so in our case 10 th term should be 2*10= 20. not one of the options.
3rd is that a1...a10 could be odd consecutive nos. in that case nth term equals to 2n-1, so in our case 10th term should be 2*10-1=19. Option D our ans.

Please correct me if my thought process is wrong.

Bunuel wrote:
nave81 wrote:
In a certain sequence of numbers, a1,a2,a3,...,an, the average (arithmetic mean) of the first m consecutive terms starting with a1 is m, for any positive integer m. If a1=1, what is a10?

A. 100
B. 55
C. 21
D. 19
E. 1

The average of the first 9 consecutive terms starting with a1 is 9 --> $$\frac{a_1+a_2+a_3+...+a_{9}}{9}=9$$ --> $$a_1+a_2+a_3+...+a_{9}=81$$.

The average of the first 10 consecutive terms starting with a1 is 10 --> $$\frac{a_1+a_2+a_3+...+a_{10}}{10}=10$$ --> $$a_1+a_2+a_3+...+a_{10}=100$$.

Subtract the first equation from the second: $$a_{10}=100-81=19$$.

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Kudos [?]: 10 [0], given: 19

Re: In a certain sequence of numbers, a1,a2,a3,...,an, the avera [#permalink]

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24 Oct 2013, 16:33
or a simpler way is this:
Since the mean of m consectuive terms is m
(a1+a10)/2=10

Solve for a10 as we know a1

1+a10=20
a10=19

Hope this helps.
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Re: In a certain sequence of numbers, a1,a2,a3,...,an, the avera [#permalink]

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24 Oct 2013, 21:00
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Expert's post
nave81 wrote:
In a certain sequence of numbers, a1,a2,a3,...,an, the average (arithmetic mean) of the first m consecutive terms starting with a1 is m, for any positive integer m. If a1=1, what is a10?

A. 100
B. 55
C. 21
D. 19
E. 1

Or use pattern recognition to solve it.
a1 = 1
a2 = ? the mean of a1 and a2 must be 2. Since a1 is 1, a2 must be 3
a2 = 3
a3 = ? the mean of a1, a2, a3 = 3. Since a1 and a2 are 1 and 3, a3 must be 5 to give a mean of 3
a3 = 5
a4 = ? the mean of a1, a2, a3 and a4 is 4. Since we have 1, 3, 5 and we need a mean of 4, a4 must be 7 (use deviation from mean to figure out each of these in a couple of secs)
a4 = 7

We see the pattern: 1, 3, 5, 7 ...
nth term is given by 2n - 1.
a10 is 2*10 - 1 = 19
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Re: In a certain sequence of numbers, a1, a2, a3, ..., an, the [#permalink]

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18 Dec 2013, 16:42
nave81 wrote:
In a certain sequence of numbers, a1, a2, a3, ..., an, the average (arithmetic mean) of the first m consecutive terms starting with a1 is m, for any positive integer m. If a1=1, what is a10?

A. 100
B. 55
C. 21
D. 19
E. 1

Sum of 'm' terms / m = Average of m

Now, average of m is also = (L + 1)/2 * (m)
Where, L is the last number of the sequence and 'm' is the number of terms
So then we have that L = 2m-1

For a10 then, we would have 2(10)-1 = 19

Hope it helps
Cheers!
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Re: In a certain sequence of numbers, a1, a2, a3, ..., an, the [#permalink]

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20 Feb 2014, 13:29
Instead of consecutive terms I would say evenly spaced set in the question. Anyway, from the example we know that they are talking about consecutive odd numbers (n^2 is the sum of the first n positive odd numbers) therefore, the largest number of the first 10 odd integers is 19.

Hope it helps
Cheers
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Re: In a certain sequence of numbers, a1, a2, a3, ..., an, the [#permalink]

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20 Feb 2014, 14:57
2 ways algebra focused and 1 way emphasizing pattern recognition:

1) by given formula:the mean of a1 is 1 the mean of a1+ a2 is 2 the mean of a1+a2+ a3 is 3 and so on. likewise for mean of numbers up toaX it is X.

if mean of numbers up to a10 is 10 then sum of numbers for a10 is 10*10. likewise, the sum of numbers for a9 is 9*9.

100 -81 is the final term: a10 =19

Formula for sum of number (( first + last)/2 )*m number of terms.

2) more elegant way:

we already know that a1 = 1 and formula for sum of numbers. ((a1 +a10)/2)*m

since average of numbers = m then sum of number = m*m

so we have an algebraic equivalency. we know a1= 1 , m= 10 thus only 1 variable to determine is a10.

3) pattern recognition

a1 = 1
a2 = ? since the mean of a1 and a2 is 2. Since a1 is 1, a2 must be 3
a2 = 3
a3 = ? since the mean of a1, a2, a3 = 3. Since a1 and a2 are 1 and 3, a3 must be 5 to give a mean of 3
a3 = 5

the pattern: 1, 3, 5,

nth term is given by 2n - 1.

a10 is 2*10 - 1 = 19

hope this helps. Thanks!
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Re: In a certain sequence of numbers, a1, a2, a3, ..., an, the [#permalink]

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In a certain sequence of numbers, a1, a2, a3, ..., an, the [#permalink]

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02 Jul 2016, 01:11
nave wrote:
In a certain sequence of numbers, a1, a2, a3, ..., an, the average (arithmetic mean) of the first m consecutive terms starting with a1 is m, for any positive integer m. If a1=1, what is a10?

A. 100
B. 55
C. 21
D. 19
E. 1

Another way..
$$a_1 = 1 (1^2 - 0^2)$$
$$a_2 = 3 (2^2 - 1^2)$$
$$a_3 = 5 (3^2 - 2^2)$$

Thus..

$$a_{10} = 19 (10^2 - 9^2)$$

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In a certain sequence of numbers, a1, a2, a3, ..., an, the   [#permalink] 02 Jul 2016, 01:11
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