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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(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:

Similar Questions

If $1, \log_9(3^{1-x} + 2), \log_3(4 \cdot 3^x - 1)$ are in $A.P.$, then $x$ equals:

If $\log _{5} 2, \log _{5}(2^{x}-3)$ and $\log _{5}(\frac{17}{2}+2^{x-1})$ are in $A.P.$,then the value of $x$ is:

The sum to infinity of the progression $9 - 3 + 1 - \frac{1}{3} + .....$ is

If $\tan n\theta = \tan m\theta$,then the different values of $\theta$ will be in

If $a, b, c$ are positive integers,then $(a + b)(b + c)(c + a)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module