In a certain sequence, the term

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

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

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

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

20*[(k+1/2)+(k+20/2)]/2=35
k=-3.5
A

term 10=k+5
term 11=k+5.5
median=k+5.25
20(k+5.25)=35→
k=-3.5
A

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