JEE Main 2024 — Sequences And Series Question with Solution

Since $3,a,b,c$ are in arithmetic progression (A.P.), the common difference can be calculated using the term $a$ (the second term) as follows:

$$d=a-3$$

The nth term of an A.P. is given by the formula:

$$T_n=a+(n-1)d$$

So, using this formula, we can express $b$ and $c$ in terms of $a$ and $d$:

$$b=a+d$$

$$c=a+2d$$

Substituting $d=a-3$ into these expressions:

$$b=a+(a-3)$$

$$c=a+2(a-3)$$

Therefore:

$$b=2a-3$$

$$c=3a-6$$

Now, let's consider that $3,a-1,b+1,c+9$ are in geometric progression (G.P.). For terms in a G.P., the ratio (common ratio, r) between consecutive terms is constant. So:

$$\frac{a-1}{3}=\frac{b+1}{a-1}=\frac{c+9}{b+1}$$

Now, we will establish the relation between the terms using the property of G.P.:

$$\frac{a-1}{3}=\frac{b+1}{a-1}$$

$$(a-1{)}^2=3(b+1)$$

$$a^2-2a+1=3b+3$$

Substituting $b=2a-3$, we get:

$$a^2-2a+1=3(2a-3)+3$$

$$a^2-2a+1=6a-9+3$$

$$a^2-8a+7=0$$

Solving this quadratic equation:

$$(a-7)(a-1)=0$$

Hence, $a=7$ or $a=1$. However, if $a=1$, the terms $3,a-1,b+1,c+9$ cannot form a G.P. as it would involve division by zero. Therefore, $a=7$. We use this value to find $b$ and $c$:

$$b=2a-3=2(7)-3=14-3=11$$

$$c=3a-6=3(7)-6=21-6=15$$

Now we can find the arithmetic mean ($A$) of $a$, $b$, and $c$:

$$A=\frac{a+b+c}{3}$$

$$A=\frac{7+11+15}{3}$$

$$A=\frac{33}{3}$$

$$A=11$$

Hence, the arithmetic mean of $a$, $b$, and $c$ is $11$, which corresponds to Option D.