Question 2

What is the sum of first 40 terms of 1 + 3 + 4 + 5 + 7 + 7 + 10 + 9 + ….?

Solution

The sequence has to be split into two with alternate terms:

S1=1+4+7+10+…(S1=1+4+7+10+…(AP with A1=1,D1=3

S2=3+5+7+9+...(S2=3+5+7+9+...(AP with A2=3,D2=2)

Sum of 40 terms of the original series = Sum of 20 terms of S1+ Sum of 20 terms of S2

S=N[2A1+(N−1)D1]/2+N[2A2+(N−1)D2]/2

⟹S=[20∗((2∗1)+(19∗3))]+[20∗((2∗3)+(19∗2)]/2

⟹S=10∗[59+44]=1030

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