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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 :

JEE · Math · previous-year question

  1. A.-4
  2. B.-1
  3. C.13
  4. D.11correct

Answer

D. 11

Explanation

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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