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In a sequence of consecutive numbers the sum of the first half of the [#permalink]
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Bunuel wrote:
In a sequence of consecutive numbers the sum of the first half of the numbers is 196. If there are 14 numbers in the set, what is the sum of the second half of the numbers?

A. 201
B. 206
C. 210
D. 245
E. 266

With 14 consecutive integers, the first half of the numbers are terms 1 through 7. Sum of these first seven = 196

For consecutive integers,
First term = n
Seventh term = n + 6

Sum of first half of numbers:

(Average)(# of terms) = Sum

Average = \(\frac{(First Term+LastTerm)}{2}\)

\(\frac{(n + n + 6)}{2} * 7 = 196\)

\((2n + 6)(7) = 392\)
\(14n + 42 = 392\)
\(14n = 350\)
\(n = 25\)

Second half of numbers = terms 8 to 14
8th term: (n+7) = (25+7) = 32
14th term: (n+13)=(25+13) = 38
Sum of second half of numbers, from 32 to 38:

\(\frac{32 + 38}{2}*(7)=(35*7)=245\)

Answer D
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Re: In a sequence of consecutive numbers the sum of the first half of the [#permalink]
chetan2u wrote:
Bunuel wrote:
In a sequence of consecutive numbers the sum of the first half of the numbers is 196. If there are 14 numbers in the set, what is the sum of the second half of the numbers?

A. 201
B. 206
C. 210
D. 245
E. 266



hi..

various ways to do it..

1) average..
consecutive numbers so average will be the middle number = \(\frac{196}{7}=28\)
average of next half = \(28+7=35\), so SUM of these numbers = \(35*7=245\)

2) Sum of each number..
first of 1st half will be 7 less than first of 2nd half
second of 1st half will be 7 less than second of 2nd half and so ontill
seventh of 1st half will be 7 less than first of 2nd half..
total of 2nd half therefore will be 7*7 more than SUM of 1st half ..
sum = 196+49=245

D


Good night chetan2u !

Could you please explain to me which is the concept behind the following that you wrote?

average of next half = \(28+7=35\)

I mean, that holds true all the time?

Kind regards!!!
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Re: In a sequence of consecutive numbers the sum of the first half of the [#permalink]
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jfranciscocuencag wrote:
chetan2u wrote:
Bunuel wrote:
In a sequence of consecutive numbers the sum of the first half of the numbers is 196. If there are 14 numbers in the set, what is the sum of the second half of the numbers?

A. 201
B. 206
C. 210
D. 245
E. 266



hi..

various ways to do it..

1) average..
consecutive numbers so average will be the middle number = \(\frac{196}{7}=28\)
average of next half = \(28+7=35\), so SUM of these numbers = \(35*7=245\)

2) Sum of each number..
first of 1st half will be 7 less than first of 2nd half
second of 1st half will be 7 less than second of 2nd half and so ontill
seventh of 1st half will be 7 less than first of 2nd half..
total of 2nd half therefore will be 7*7 more than SUM of 1st half ..
sum = 196+49=245

D


Good night chetan2u !

Could you please explain to me which is the concept behind the following that you wrote?

average of next half = \(28+7=35\)

I mean, that holds true all the time?

Kind regards!!!


Yes,
Since the numbers are consecutive..
Say first 7 are (a-3),(a-2),(a-1),a,(a+1),(a+2),(a+3).. a is the mean and median..
Next 7 in continuity will be (a+4),(a+5),(a+6),(a+7),(a+8),(a+9),(a+10). Here a+7 is the median and mean
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Re: In a sequence of consecutive numbers the sum of the first half of the [#permalink]
Bunuel wrote:
In a sequence of consecutive numbers the sum of the first half of the numbers is 196. If there are 14 numbers in the set, what is the sum of the second half of the numbers?

A. 201
B. 206
C. 210
D. 245
E. 266


Seven consecutive numbers will be

x x+1 x+2 x+3 x+4 x+5 x+6, now the next seven will be
x+7 x+8 x+9 x+10 x+11 x+12 x+13,
the difference between the 1st term and the 8th term will be of 7, which will hold good for the rest of the numbers as well

So 7*7 = 49

196 + 49
245

D
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Re: In a sequence of consecutive numbers the sum of the first half of the [#permalink]
Bunuel wrote:
In a sequence of consecutive numbers the sum of the first half of the numbers is 196. If there are 14 numbers in the set, what is the sum of the second half of the numbers?

A. 201
B. 206
C. 210
D. 245
E. 266


7 consective integers sum = 7a+21= 196
a= 25
so 8th term in sequence ; a+7 = 25+7 = 32
and 14th term = a+13 = 25+13 = 38
sum of terms 32 to 38
sn = n/2 * ( 2a+(n-1) *d)
n=7
a=32
d=1
s= 7/2 * ( 32*2 +6)
s= 245
IMO D
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Re: In a sequence of consecutive numbers the sum of the first half of the [#permalink]
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Bunuel wrote:
In a sequence of consecutive numbers the sum of the first half of the numbers is 196. If there are 14 numbers in the set, what is the sum of the second half of the numbers?

A. 201
B. 206
C. 210
D. 245
E. 266


After looking at how other people solved the question, I would not recommend my method...

\(\frac{7}{2} * (2a + 6) = 196\)

Solve for first term and we get 25

This means the next half of our sequence begins at 32.

\(\frac{7}{2} * (2*32 + 6) = 245\)
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Re: In a sequence of consecutive numbers the sum of the first half of the [#permalink]
kchen1994 wrote:
Bunuel wrote:
In a sequence of consecutive numbers the sum of the first half of the numbers is 196. If there are 14 numbers in the set, what is the sum of the second half of the numbers?

A. 201
B. 206
C. 210
D. 245
E. 266


After looking at how other people solved the question, I would not recommend my method...

\(\frac{7}{2} * (2a + 6) = 196\)

Solve for first term and we get 25

This means the next half of our sequence begins at 32.

\(\frac{7}{2} * (2*32 + 6) = 245\)


kchen1994

If this approach works for you, in other questions of this type, thereby, enabling you to solve those questions within 2 minutes.

You should follow this. :-)

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