If the sum of the first 40 terms of the series 3+4+8+9+13+14+18+19+ is (102)m, then m is e…

Question

If the sum of the first 40 terms of the series $3+4+8+9+13+14+18+19+\cdots$ is $(102)m$, then $m$ is equal to:

Accepted Answer

Grouping in pairs: $(3+4)+(8+9)+(13+14)+\cdots = 7+17+27+37+\cdots$ (20 terms, A.P. common difference $10$).

$$S_{40} = \frac{20}{2}\big[2(7)+19(10)\big] = 10(204) = 2040 = 102\times20.$$

So $m=20$.

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