Question

the sum if the series 1/2 + 1/3 + 1/6 + ... to 9 terms is​

Answer 1

Answer:

Given,

a=1/2=a1

a2=1/3

d=a2‐a1=1/3‐1/2=2‐3/6=‐1/6

n=9

Sn=n/2[2a+(n‐1)d]

S9=9/2[2(1/2)+(9‐1)(‐1/6)]

S9=9/2[1‐8/6]

S9=9/2[‐2/6]

S9=‐3/2

Answer 2

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