If the sum of the first 40 terms of the serie | vedclass

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(A) The given series is $3+4+8+9+13+14+18+19+\ldots$ up to $40$ terms.This can be split into two arithmetic progressions,each having $20$ terms:Series

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$20$

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(A) The given series is $3+4+8+9+13+14+18+19+\ldots$ up to $40$ terms.This can be split into two arithmetic progressions,each having $20$ terms:Series $1$: $3, 8, 13, 18, \ldots$ ($20$ terms)Series $2$: $4, 9, 14, 19, \ldots$ ($20$ terms)Sum of Series $1$ $(S_1)$ = $\frac{20}{2} [2(3) + (20-1)5] = 10 [6 + 95] = 10 	imes 101 = 1010$.Sum of Series $2$ $(S_2)$ = $\frac{20}{2} [2(4) + (20-1)5] = 10 [8 + 95] = 10 	imes 103 = 1030$.Total Sum = $S_1 + S_2 = 1010 + 1030 = 2040$.Given that the total sum is $(102)m$,we have $2040 = 102m$.Therefore,$m = \frac{2040}{102} = 20$.

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

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