Find the sum of all numbers between 200 and 4 | vedclass

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

Question

(A) The numbers between $200$ and $400$ that are divisible by $7$ form an arithmetic progression: $203, 210, 217, \dots, 399.$Here,the first term $a =

Vedclass · Accepted Answer

$8729$

Vedclass · Answer

(A) The numbers between $200$ and $400$ that are divisible by $7$ form an arithmetic progression: $203, 210, 217, \dots, 399.$Here,the first term $a = 203$ and the last term $l = 399.$The common difference $d = 7.$Using the formula for the $n^{th}$ term of an $A.P.$,$a_n = a + (n - 1)d:$$399 = 203 + (n - 1)7$$196 = (n - 1)7$$n - 1 = 28$$n = 29.$The sum of $n$ terms is given by $S_n = \frac{n}{2}(a + l):$$S_{29} = \frac{29}{2}(203 + 399)$$S_{29} = \frac{29}{2}(602)$$S_{29} = 29 	imes 301 = 8729.$Thus,the required sum is $8729.$

Find the sum of all numbers between $200$ and $400$ which are divisible by $7.$

Similar Questions

The number of terms common to the two $A$.$P$.'s $3, 7, 11, \ldots, 407$ and $2, 9, 16, \ldots, 709$ is

If $x, y, z \in \mathbb{R}^+$ such that $x + y + z = 4$,then the maximum possible value of $xyz^2$ is

If the $19^{th}$ term of a non-zero $A.P.$ is zero,then the ratio of its ($49^{th}$ term) to ($29^{th}$ term) is:

The maximum sum of the series $20 + 19\frac{1}{3} + 18\frac{2}{3} + \dots$ is

If the roots $\alpha, \beta, \gamma$ of the equation $x^3-6x^2+px+10=0$ are in arithmetic progression,then $\alpha^3+\beta^3+\gamma^3=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module