Aptitude | Progressions | Question 5

Find sum of series: 2, 2.5, 3, 3. 5, 4, 4. 5……….11

(A)

120

(B)

123.5

(C)

126.5

(D)

118.5

Answer: (B)

Explanation:

To, find the number of terms in the Ap 
Tn = a+(n-1)*d
11 = 2+(n-1)*0.5
So, we get n = 19

sum of AP = (n/2)[2a+(n-1)d]
n=19, a=2, d=1/2
S = (19/2)[2*2+(19-1)1/2]
=(19/2)[4+9] 
=9.5*13 = 123.5

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