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# In a certain sequence, the term

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CEO
Joined: 11 Sep 2015
Posts: 3122
In a certain sequence, the term  [#permalink]

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26 Jan 2017, 11:28
Top Contributor
5
00:00

Difficulty:

35% (medium)

Question Stats:

72% (02:47) correct 28% (02:49) wrong based on 193 sessions

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In a certain sequence, the term an is given by the formula an = k + $$\frac{n}{2}$$, where k is a constant. If the sum of all the terms from a1 to a20 inclusive equals 35, what is the value of k?

A) -3.5
B) - 3.2
C) -3
D) -2.8
E) -2.5

*Kudos for all correct solutions

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CEO
Joined: 11 Sep 2015
Posts: 3122
Re: In a certain sequence, the term  [#permalink]

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27 Jan 2017, 08:09
2
Top Contributor
4
GMATPrepNow wrote:
In a certain sequence, the term an is given by the formula an = k + $$\frac{n}{2}$$, where k is a constant. If the sum of all the terms from a1 to a20 inclusive equals 35, what is the value of k?

A) -3.5
B) - 3.2
C) -3
D) -2.8
E) -2.5

Let's examine a few terms:
a1 = k + 1/2
a2 = k + 2/2
a3 = k + 3/2
.
.
.
a20 = k + 20/2

So, TOTAL SUM = (k + 1/2) + (k + 2/2) + (k + 3/2) + . . . + (k + 20/2)
= 20k + 1/2 + 2/2 + 3/2 + . . . + 20/2
= 20k + (1/2)(1 + 2 + 3 + . . . + 20)

Sum of first n integers formula: 1 + 2 + 3 + 4 + . . . + n = (n)(n + 1)/2

Applying this formula to the sum 1 + 2 + 3 + . . . + 20, we get:
= 20k + (1/2)[(20)(21)/2 ]
= 20k + 105

We're told that this sum equals 35, so......
35 = 20k + 105
Subtract 105 from both sides: -70 = 20k
Divide both sides by 20 to get: -3.5 = k

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##### General Discussion
Manager
Joined: 03 Oct 2013
Posts: 84
Re: In a certain sequence, the term  [#permalink]

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26 Jan 2017, 11:37
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Retired Moderator
Joined: 27 May 2014
Posts: 525
GMAT 1: 730 Q49 V41
Re: In a certain sequence, the term  [#permalink]

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26 Jan 2017, 18:50
2
Given that An = K + n/2;

Therefore, A1 = K + 1/2; A2 = K + 2/2 ..... A20 = K + 20/2

Sum of A1 to A20 = 20K + 1/2( 1 + 2 + 3 + .... 19 + 20)
=>35 = 20K + 1/2*20 (1+20)/2
=>35 = 20K + 105
=>20K = -70
=>K = -3.5. Ans A
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Joined: 22 Nov 2016
Posts: 213
Location: United States
GPA: 3.4
Re: In a certain sequence, the term  [#permalink]

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01 Jul 2017, 11:29
1
1
Writing down first few numbers of the sequence...

$$a(1)= k+\frac{1}{2}$$
$$a(2)= k+\frac{2}{2}$$
$$a(3)= k+\frac{3}{2}$$
.
.
.

The sum of $$35=20k+\frac{1}{2}*(1+2+....+19+20)$$
$$35 = 20k+\frac{(20*21)}{4}$$
$$35 = 20K+105$$
$$20k=-70$$
$$k=\frac{-7}{2}$$ = -3.5, A
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Intern
Joined: 14 Oct 2016
Posts: 30
Location: India
WE: Sales (Energy and Utilities)
Re: In a certain sequence, the term  [#permalink]

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08 Sep 2017, 19:07
1
We can find the answer in two methods

First Method:
a(1)= K+1/2, a(2)= K+1 , a(3)=K=3/2 then we see each term is + 1/2 the previous term.

For AP
Sum = n{(2a+(n-1)d}/2
So 35= 20{2a+ (19)*1/2)}/2
Simplifying we get
7=2a+19
So a= -3,
so a(1)= -3= K+1/2
So k= -7/2= -3.5

Second Method:
So each even term will be consecutive integer 1,2,3,4,5,6,7,8,9,10 and each odd term will be fraction with numerators as odd consecutive numbers and denominator as 2
like 1/2, 3/2,5/2,7/2.9/2,11/2,13/2,15/2,17/2,19/2

So the sum will be
35= 20 K + {1+3+5+7+9+11+13+15+17+19}/2 + {1+2+3+4+5+6+7+8+9+10)

Formula for sum of n odd integers where n is odd is {(n+1)/2}^2 and formula for sum of n odd integers where n is even{(n/2)}^2

Say we have to find sum of 19 odd integers . In this case we use formula {(n+1)/2}^2 since n=19 is odd

And formula for sum of n natural numbers is n(n+1)/1

35= 20K +{(19+1)/2}^2+ 10(11)/2

35= 20k +55+50
k= -3.5
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Abhimanyu

Intern
Joined: 17 Nov 2016
Posts: 26
Re: In a certain sequence, the term  [#permalink]

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06 Mar 2018, 16:42
What is wrong with this approach?

A1= K + 1/2
A20= K +20/2 = K + 10

Sum of the sequence is: ((1st term + Last term)/2) * n (n is number of terms)
35 = ((A1 + A2)/2) * 20
35 = (K +1/2 + K+10)/2 * 20
35 = (5k/2 + 10) * 10
35 = 25K + 100
25K = -65
K = -13/5
K = -2.6

What went wrong here?
CEO
Joined: 11 Sep 2015
Posts: 3122
In a certain sequence, the term  [#permalink]

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Updated on: 09 May 2018, 05:55
Top Contributor
Zoser wrote:
What is wrong with this approach?

A1= K + 1/2
A20= K +20/2 = K + 10

Sum of the sequence is: ((1st term + Last term)/2) * n (n is number of terms)
35 = ((A1 + A2)/2) * 20
35 = (K +1/2 + K+10)/2 * 20
35 = (5k/2 + 10) * 10
35 = 25K + 100
25K = -65
K = -13/5
K = -2.6

What went wrong here?

Everything is fine up to: 35 = (K +1/2 + K+10)/2 * 20
Simplify to get: 35 = (2K + 10.5)/2 * 20
Multiply both sides by 2 to get: 70 = (2K + 10.5)(20)
Expand right side: 70 = 40K + 210
Subtract 210 from both sides: -140 = 40k
Solve: k = -140/40 = -14/4 = -7/2 = -3.5

Cheers,
Brent
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Originally posted by GMATPrepNow on 08 May 2018, 12:13.
Last edited by GMATPrepNow on 09 May 2018, 05:55, edited 1 time in total.
VP
Joined: 07 Dec 2014
Posts: 1116
Re: In a certain sequence, the term  [#permalink]

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08 May 2018, 22:19
1
GMATPrepNow wrote:
In a certain sequence, the term an is given by the formula an = k + $$\frac{n}{2}$$, where k is a constant. If the sum of all the terms from a1 to a20 inclusive equals 35, what is the value of k?

A) -3.5
B) - 3.2
C) -3
D) -2.8
E) -2.5

*Kudos for all correct solutions

20*[(k+1/2)+(k+20/2)]/2=35
k=-3.5
A
Re: In a certain sequence, the term &nbs [#permalink] 08 May 2018, 22:19
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