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NVAMediumJEE 2023Geometric Progression (GP)

JEE Mathematics 2023 Question with Solution

For the two positive numbers a,ba, b, if a,ba, b and 118\frac{1}{18} are in a geometric progression, while 1a,10,1b\frac{1}{a}, 10, \frac{1}{b} are in an arithmetic progression, then 16a+12b16a + 12b is equal to:

Answer

Correct answer:3

Step-by-step solution

Standard Method

Given: a,b,118a, b, \frac{1}{18} are in GP and 1a,10,1b\frac{1}{a}, 10, \frac{1}{b} are in AP.

Find: 16a+12b16a + 12b.

From the GP condition,

a18=b2\frac{a}{18} = b^2

So,

a=18b2a = 18b^2

From the AP condition,

1a+1b=20\frac{1}{a} + \frac{1}{b} = 20

Multiplying by abab,

a+b=20aba + b = 20ab

Now substitute a=18b2a = 18b^2:

18b2+b=20⋅18b2⋅b18b^2 + b = 20 \cdot 18b^2 \cdot b 18b2+b=360b318b^2 + b = 360b^3

Since b>0b > 0, divide by bb:

18b+1=360b218b + 1 = 360b^2 360b2−18b−1=0360b^2 - 18b - 1 = 0

Solving this quadratic gives

b=112b = \frac{1}{12}

Using a=18b2a = 18b^2,

a=18(112)2=18a = 18 \left(\frac{1}{12}\right)^2 = \frac{1}{8}

Now,

16a+12b=16×18+12×112=2+1=316a + 12b = 16 \times \frac{1}{8} + 12 \times \frac{1}{12} = 2 + 1 = 3

Therefore, the required value is 33.

Direct Substitution Route

Given: a,b,118a, b, \frac{1}{18} are in GP and 1a,10,1b\frac{1}{a}, 10, \frac{1}{b} are in AP.

Find: 16a+12b16a + 12b.

Use the two progression conditions directly:

a=18b2a = 18b^2

and

a+b=20aba + b = 20ab

Substituting a=18b2a = 18b^2 into the second relation,

18b2+b=360b318b^2 + b = 360b^3

This leads to

360b2−18b−1=0360b^2 - 18b - 1 = 0

Taking the positive root,

b=112b = \frac{1}{12}

Then

a=18⋅1144=18a = 18 \cdot \frac{1}{144} = \frac{1}{8}

Hence,

16a+12b=16⋅18+12⋅112=316a + 12b = 16 \cdot \frac{1}{8} + 12 \cdot \frac{1}{12} = 3

So the answer is 33.

Common mistakes

  • Using the GP condition incorrectly. In a geometric progression, the square of the middle term equals the product of the extremes, so b2=a⋅118b^2 = a \cdot \frac{1}{18}. Do not write relations like ab=118ab = \frac{1}{18}. Use a18=b2\frac{a}{18} = b^2.

  • Applying the AP condition wrongly. For three terms in AP, the middle term is the average of the other two, so 10=12(1a+1b)10 = \frac{1}{2}\left(\frac{1}{a} + \frac{1}{b}\right), which gives 1a+1b=20\frac{1}{a} + \frac{1}{b} = 20. Do not equate consecutive differences without care.

  • Ignoring the condition that the numbers are positive. The quadratic in bb gives two roots, but only the positive value is valid here. Always check the sign restriction before choosing the final root.

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