Let 3, a, b, c be in A.P.

Question

Let $3, a, b, c$ be in A.P. and $3, a-1, b+1, c+9$ be in G.P. Then, the arithmetic mean of $a, b$ and $c$ is :

Accepted Answer

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.

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