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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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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440 to 730: If I Can, You Can Too

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

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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