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In a certain sequence, the term

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In a certain sequence, the term [#permalink]

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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

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[Reveal] Spoiler: OA

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The answer is A. Please see attached image for calculations.
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Untitled.png [ 81.23 KiB | Viewed 943 times ]


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Re: In a certain sequence, the term [#permalink]

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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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New post 27 Jan 2017, 09:09
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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

Answer: A
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Re: In a certain sequence, the term [#permalink]

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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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Re: In a certain sequence, the term [#permalink]

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New post 08 Sep 2017, 20:07
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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Re: In a certain sequence, the term   [#permalink] 08 Sep 2017, 20:07
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