JEE Mathematics 2023 Question with Solution

Given: $A_1$ and $A_2$ are two arithmetic means, and $G_1, G_2, G_3$ are three geometric means between two distinct positive numbers.

Find: The value of $G_1^{4}+G_2^{4}+G_3^{4}+G_1^{2}G_3^{2}$ in terms of $A_1, A_2, G_1, G_3$.

Let the two distinct positive numbers be $a$ and $b$.

If $G_1, G_2, G_3$ are three geometric means between $a$ and $b$, then

$$a,\,G_1,\,G_2,\,G_3,\,b$$

form a GP. Hence,

$$G_1=ar,\quad G_2=ar^2,\quad G_3=ar^3,\quad b=ar^4$$

The two arithmetic means between $a$ and $b$ are

$$A_1=\frac{2a+b}{3},\quad A_2=\frac{a+2b}{3}$$

Thus,

$$A_1+A_2=a+b$$

Now evaluate the given expression:

$$G_1^{4}+G_2^{4}+G_3^{4}+G_1^{2}G_3^{2}$$ $$=(ar)^4+(ar^2)^4+(ar^3)^4+(ar)^2(ar^3)^2$$ $$=a^4(r^4+r^8+r^{12}+r^8)$$ $$=a^4(r^4+2r^8+r^{12})$$ $$=a^4(r^2+r^6)^2$$

Since $b=ar^4$, we have

$$a+b=a(1+r^4)$$

and

$$G_1G_3=(ar)(ar^3)=a^2r^4$$

Hence,

$$(a+b)^2G_1G_3=a^2(1+r^4)^2\cdot a^2r^4=a^4(r^4+2r^8+r^{12})$$

This matches the given expression. Therefore,

$$G_1^{4}+G_2^{4}+G_3^{4}+G_1^{2}G_3^{2}=(A_1+A_2)^2G_1G_3$$

The correct option is C.

Given: Three geometric means lie between two positive numbers.

Find: A compact way to rewrite the expression.

Write the numbers in GP form as $a, ar, ar^2, ar^3, ar^4$. Then

$$G_1=ar,\quad G_2=ar^2,\quad G_3=ar^3$$

and

$$A_1+A_2=a+b=a+ar^4=a(1+r^4)$$

Also,

$$G_1G_3=a^2r^4$$

Now,

$$G_1^{4}+G_2^{4}+G_3^{4}+G_1^{2}G_3^{2}$$ $$=a^4(r^4+r^8+r^{12}+r^8)$$ $$=a^4(r^4+2r^8+r^{12})$$ $$=a^4r^4(1+r^4)^2$$

But

$$(a+b)^2G_1G_3=a^2(1+r^4)^2\cdot a^2r^4=a^4r^4(1+r^4)^2$$

So the expression is equal to $(A_1+A_2)^2G_1G_3$. The correct option is C.