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