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MCQMediumJEE 2023Measures of Dispersion

JEE Mathematics 2023 Question with Solution

Let the six numbers a1,a2,a3,a4,a5,a6a_1, a_2, a_3, a_4, a_5, a_6 be in A.P., and a1+a3=10a_1 + a_3 = 10. If the mean of these six numbers is 192\frac{19}{2} and their variance is σ2\sigma^2, then 8σ28\sigma^2 is equal to:

  • A

    220220

  • B

    210210

  • C

    200200

  • D

    105105

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The six numbers are in A.P. and satisfy a1+a3=10a_1 + a_3 = 10. Their mean is 192\frac{19}{2}.

Find: The value of 8σ28\sigma^2.

Write the six terms as

a1, a1+d, a1+2d, a1+3d, a1+4d, a1+5da_1,\ a_1+d,\ a_1+2d,\ a_1+3d,\ a_1+4d,\ a_1+5d

Using a1+a3=10a_1+a_3=10,

a1+(a1+2d)=10a_1+(a_1+2d)=10 2a1+2d=102a_1+2d=10 a1+d=5a_1+d=5

Now use the mean of the six numbers:

a1+(a1+d)+(a1+2d)+(a1+3d)+(a1+4d)+(a1+5d)6=192\frac{a_1+(a_1+d)+(a_1+2d)+(a_1+3d)+(a_1+4d)+(a_1+5d)}{6}=\frac{19}{2} 6a1+15d6=192\frac{6a_1+15d}{6}=\frac{19}{2} a1+5d2=192a_1+\frac{5d}{2}=\frac{19}{2} 2a1+5d=192a_1+5d=19

Solve the two equations

a1+d=5a_1+d=5

and

2a1+5d=192a_1+5d=19

From a1=5−da_1=5-d, substitute into the second equation:

2(5−d)+5d=192(5-d)+5d=19 10−2d+5d=1910-2d+5d=19 3d=93d=9 d=3d=3

Then

a1=5−3=2a_1=5-3=2

So the six numbers are

2, 5, 8, 11, 14, 172,\ 5,\ 8,\ 11,\ 14,\ 17

Use variance formula:

σ2=mean of squares−(mean)2\sigma^2=\text{mean of squares}-(\text{mean})^2

Hence

σ2=22+52+82+112+142+1726−(192)2\sigma^2=\frac{2^2+5^2+8^2+11^2+14^2+17^2}{6}-\left(\frac{19}{2}\right)^2 σ2=4+25+64+121+196+2896−3614\sigma^2=\frac{4+25+64+121+196+289}{6}-\frac{361}{4} σ2=6996−3614\sigma^2=\frac{699}{6}-\frac{361}{4} σ2=116.5−90.25=26.25\sigma^2=116.5-90.25=26.25

Therefore,

8σ2=8×26.25=2108\sigma^2=8\times 26.25=210

The correct option is B.

Equation-by-Equation Derivation

Given: Six numbers in A.P. with a1+a3=10a_1+a_3=10 and mean 192\frac{19}{2}.

Find: 8σ28\sigma^2.

Since the sequence is in A.P., the terms are

a1, a2, a3, a4, a5, a6a_1,\ a_2,\ a_3,\ a_4,\ a_5,\ a_6

with

a2=a1+d,a3=a1+2d,a4=a1+3d,a5=a1+4d,a6=a1+5da_2=a_1+d,\quad a_3=a_1+2d,\quad a_4=a_1+3d,\quad a_5=a_1+4d,\quad a_6=a_1+5d

The condition a1+a3=10a_1+a_3=10 gives

a1+(a1+2d)=10a_1+(a_1+2d)=10 2a1+2d=102a_1+2d=10 a1+d=5a_1+d=5

The mean condition gives total sum

6×192=576\times \frac{19}{2}=57

Also, sum of the six terms is

a1+(a1+d)+(a1+2d)+(a1+3d)+(a1+4d)+(a1+5d)=6a1+15da_1+(a_1+d)+(a_1+2d)+(a_1+3d)+(a_1+4d)+(a_1+5d)=6a_1+15d

So

6a1+15d=576a_1+15d=57 2a1+5d=192a_1+5d=19

Now subtract twice the first relation from the second form carefully:

2a1+5d=192a_1+5d=19 2a1+2d=102a_1+2d=10

Subtracting,

3d=93d=9 d=3d=3

Then from a1+d=5a_1+d=5,

a1=2a_1=2

Hence the numbers are

2, 5, 8, 11, 14, 172,\ 5,\ 8,\ 11,\ 14,\ 17

Their mean is indeed

2+5+8+11+14+176=576=192\frac{2+5+8+11+14+17}{6}=\frac{57}{6}=\frac{19}{2}

Now compute the mean of squares:

22+52+82+112+142+1726=6996\frac{2^2+5^2+8^2+11^2+14^2+17^2}{6}=\frac{699}{6}

And square of mean:

(192)2=3614\left(\frac{19}{2}\right)^2=\frac{361}{4}

Therefore

σ2=6996−3614=1054\sigma^2=\frac{699}{6}-\frac{361}{4}=\frac{105}{4}

Finally,

8σ2=8×1054=2108\sigma^2=8\times \frac{105}{4}=210

So the correct option is B.

Common mistakes

  • Using the wrong third term as a1+da_1+d instead of a1+2da_1+2d. In an A.P., the terms increase by the common difference successively. Write all six terms explicitly before applying the condition a1+a3=10a_1+a_3=10.

  • Confusing mean with sum. The mean is 192\frac{19}{2}, so the sum of six numbers is 6×192=576\times \frac{19}{2}=57, not 192\frac{19}{2} itself. Always convert mean to total sum when forming an equation.

  • Using an incorrect variance formula such as averaging deviations without squaring. Here variance is σ2=mean of squares−(mean)2\sigma^2=\text{mean of squares}-(\text{mean})^2. Compute both quantities separately and then subtract.

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