JEE Main 2023 — Sequences And Series Question with Solution

Now, we have the following relations :

Arithmetic progression :

Since $A_1$ and $A_2$ are arithmetic means between $a$ and $b$, we can say that $a$, $A_1$, $A_2$, and $b$ are in an arithmetic progression. This means there are three equal intervals between $a$ and $b$, which are represented by the common difference $d$.

To find the value of $d$, we can use the following equation :

$$b-a=3d$$

From this equation, we can find the value of $d$ :

$$d=\frac{b-a}{3}$$

$$A_1=a+\frac{b-a}{3}=\frac{2a+b}{3}$$

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

$$A_1+A_2=a+b$$

Geometric progression :

$$a,G_1,G_2,G_3,b\text{ are in G.P. }$$

$$r={\left(\frac{b}{a}\right)}^{\frac{1}{4}}$$

$$G_1={\left(a^{3}b\right)}^{\frac{1}{4}}$$

$$G_2={\left(a^2{b}^2\right)}^{\frac{1}{4}}$$

$$G_3={\left(a{b}^3\right)}^{\frac{1}{4}}$$

We have the expression :

$$G_1^4+G_2^4+G_3^4+G_1^2{G}_3^2=a^{3}b+a^2{b}^2+a{b}^3+{\left(a^{3}b\right)}^{\frac{1}{2}}\cdot {\left(a{b}^3\right)}^{\frac{1}{2}}$$

Simplify the expression :

$$a^{3}b+a^2{b}^2+a{b}^3+ab(a^2{b}^2)$$

Factor out $ab$:

$$ab(a^2+ab+b^2+a^2{b}^2)$$

Combine the terms :

$$ab(a^2+2ab+b^2)$$

Rewrite the expression using the sum of squares :

$$ab(a+b{)}^2$$

Now, recall that $A_1+A_2=a+b$. Substitute this into the expression :

$$G_1\cdot G_3\cdot (A_1+A_2{)}^2$$