If the sum of the series 2 + 5 + 8 + 11 + dot | vedclass

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(B) The given series is an Arithmetic Progression $(A.P.)$ where the first term $a = 2$ and the common difference $d = 5 - 2 = 3$.Let the number of te

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$200$

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(B) The given series is an Arithmetic Progression $(A.P.)$ where the first term $a = 2$ and the common difference $d = 5 - 2 = 3$.Let the number of terms be $n$.The sum of the first $n$ terms of an $A.P.$ is given by the formula: $S_n = \frac{n}{2} [2a + (n - 1)d]$.Given $S_n = 60100$, we have:$60100 = \frac{n}{2} [2(2) + (n - 1)3]$$120200 = n [4 + 3n - 3]$$120200 = n(3n + 1)$$3n^2 + n - 120200 = 0$Solving the quadratic equation using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$n = \frac{-1 \pm \sqrt{1^2 - 4(3)(-120200)}}{2(3)}$$n = \frac{-1 \pm \sqrt{1 + 1442400}}{6}$$n = \frac{-1 \pm \sqrt{1442401}}{6}$$n = \frac{-1 \pm 1201}{6}$Since $n$ must be positive, $n = \frac{1200}{6} = 200$.Thus, the number of terms is $200$.

If the sum of the series $2 + 5 + 8 + 11 + \dots$ is $60100$, then the number of terms is:

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