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 [2024]

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Solution

Q.
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 [2024]

1 $13$

2 $- 4$

3 $- 1$

4 $11$

Ans.
(4)

Given that 3, $a, b, c,$ are in A.P.

So, $a - 3 = b - a$

$\Rightarrow 2 a = b + 3$ ...(i)

and $b - a = c - b$

$\Rightarrow 2 b = a + c$ ...(ii)

Also, given that 3, $a - 1, b + 1, c + 9$ are in G.P.

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

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

$\Rightarrow a^{2} - 2 a = 3 (2 a - 3) + 2$ (Using (i))

$\Rightarrow a^{2} - 8 a + 7 = 0 \Rightarrow a^{2} - 7 a - a + 7 = 0$

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

$\Rightarrow (a - 1) (a - 7) = 0 \Rightarrow a = 1, 7$

By (i), $b = - 1$ and $b = 11$

Since, $b$ cannot be negative.

By (ii), $c = 15$

$∴$ A.M. of $a, b, c = \frac{a + b + c}{3} = \frac{15 + 11 + 7}{3} = \frac{33}{3} = 11$

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