The median of all 4-digit numbers that are di | vedclass

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

Question

(B) The $4$-digit numbers divisible by $7$ form an arithmetic progression: $1001, 1008, 1015, \ldots, 9996$.Here,the first term $a = 1001$ and the com

Vedclass · Accepted Answer

$5498.5$

Vedclass · Answer

(B) The $4$-digit numbers divisible by $7$ form an arithmetic progression: $1001, 1008, 1015, \ldots, 9996$.Here,the first term $a = 1001$ and the common difference $d = 7$.The last term $l = a + (n-1)d$,so $9996 = 1001 + (n-1)7$.$8995 = (n-1)7 \implies n-1 = 1285 \implies n = 1286$.Since the number of terms $n = 1286$ is even,the median is the average of the $\left(\frac{n}{2}ight)^{	ext{th}}$ and $\left(\frac{n}{2}+1ight)^{	ext{th}}$ terms.Median $= \frac{a_{643} + a_{644}}{2}$.$a_{643} = 1001 + (643-1)7 = 1001 + 4494 = 5495$.$a_{644} = 1001 + (644-1)7 = 1001 + 4501 = 5502$.Median $= \frac{5495 + 5502}{2} = \frac{10997}{2} = 5498.5$.

The median of all $4$-digit numbers that are divisible by $7$ is

Similar Questions

If $a, b, c$ are in Arithmetic Progression,then $(a - c)^2 = \dots$

Let $A_1, A_2, \dots, A_{39}$ be $39$ arithmetic means between the numbers $59$ and $159$. Then the mean of $A_{25}, A_{28}, A_{31}$ and $A_{36}$ is equal to:

Find the sum of integers from $1$ to $100$ that are divisible by $2$ or $5.$

If $a_m$ denotes the $m^{th}$ term of an $A.P.$,then $a_m$ =

If the $9^{th}$ term of an $A.P.$ is $35$ and the $19^{th}$ term is $75$,then its $20^{th}$ term will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module