The mean deviation from the mean of the A.P. is

Explanation: Step 1: Mean ( ) calculate karein A.P. ka mean hamesha uski middle term hoti hai agar terms odd hain. Step 2: Deviations calculate karein Hum har term ko mean se subtract karenge: ... ... Hamare paas deviations ki ek series banegi: Step 3: Sum of Deviations ( ) Zero ko chhod kar, har term do baar aa rahi hai (ek baar left side se, ek baar right side se): Sum of first natural numbers ka formula use karte hue: Step 4: Mean Deviation (M.D.) Formula: Yahan hai: Final Result Diye gaye A.P. ka mean deviation hai:

How do I solve CUET PG 2023 Mathematics questions?

Practice CUET PG 2023 Mathematics previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

What is the difficulty level of CUET PG Mathematics questions?

CUET PG Mathematics questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Are CUET PG previous year papers available for free?

Yes. Mathem Solvex provides free access to CUET PG previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

Tip:A–D to answerE for explanationV for videoS to reveal answer

CUET PG 2023 — Mathematics PYQ

CUET PG | Mathematics | 2023

The mean deviation from the mean of the A.P. $a,, a + d,, a + 2 d,, \dots,, a + 2 n d$ is

Choose the correct answer:

B.
$\frac{n ( n + 1 ) d}{2 n + 1}$
(Correct Answer)

C.
$\frac{n ( n + 1 ) d}{2 n}$

D.
$\frac{n ( n - 1 ) d}{2 n + 1}$

Explanation

Step 1: Mean ( $\bar{x}$ ) calculate karein

A.P. ka mean hamesha uski middle term hoti hai agar terms odd hain.

$\bar{x} = \frac{\text{First term} + \text{Last term}}{2}$

$\bar{x} = \frac{a + (a + 2nd)}{2} = \frac{2a + 2nd}{2} = a + nd$

Step 2: Deviations $|x_i - \bar{x}|$ calculate karein

Hum har term ko mean $(a+nd)$ se subtract karenge:

$|a - (a+nd)| = |-nd| = nd$

$|(a+d) - (a+nd)| = |(1-n)d| = (n-1)d$

...

$|(a+nd) - (a+nd)| = 0$

...

$|(a+2nd) - (a+nd)| = |nd| = nd$

Hamare paas deviations ki ek series banegi:

$nd, (n-1)d, \dots, d, 0, d, \dots, (n-1)d, nd$

Step 3: Sum of Deviations ( $\sum |x_i - \bar{x}|$ )

Zero ko chhod kar, har term do baar aa rahi hai (ek baar left side se, ek baar right side se):

$\sum |x_i - \bar{x}| = 2[d + 2d + 3d + \dots + nd]$

Sum of first $n$ natural numbers ka formula use karte hue:

$\sum |x_i - \bar{x}| = 2d \left[ \frac{n(n+1)}{2} \right] = n(n+1)d$

Step 4: Mean Deviation (M.D.)

Formula: $M.D. = \frac{\sum |x_i - \bar{x}|}{N}$

Yahan $N = 2n + 1$ hai:

$M.D. = \frac{n(n+1)d}{2n+1}$

Final Result

Diye gaye A.P. ka mean deviation hai:

$\frac{n(n+1)d}{2n+1}$