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Let the six numbers $$\mathrm{a_1,a_2,a_3,a_4,a_5,a_6}$$, be in A.P. and $$\mathrm{a_1+a_3=10}$$. If the mean of these six numbers is $$\frac{19}{2}$$ and their variance is $$\sigma^2$$, then 8$$\sigma^2$$ is equal to :

JEE · Math · previous-year question

  1. A.220
  2. B.210correct
  3. C.105
  4. D.200

Answer

B. 210

Explanation

$${a_1},{a_2},{a_3},{a_4},{a_5},{a_6}$$ are in AP. Let $${a_1} = a$$ $${a_2} = a + d$$ $${a_3} = a + 2d$$ $${a_4} = a + 3d$$ $${a_5} = a + 4d$$ $${a_6} = a + 5d$$ Now Mean of $${a_1},{a_2},{a_3},{a_4},{a_5}$$ and $${a_6}$$ is $$ = {{{a_1} + {a_2} + {a_3} + {a_4} + {a_5} + {a_6}} \over 6} = {{19} \over 2}$$ $$ \Rightarrow {{6a + 15d} \over 6} = {{19} \over 2}$$ $$ \Rightarrow 2a + 5d = 19$$ ...... (1) Also, given, $${a_1} + {a_3} = 10$$ $$ \Rightarrow a + a + 2d = 10$$ $$ \Rightarrow 2a + 2d = 10$$ $$ \Rightarrow a + d = 5$$ ..... (2) From equation (1) and (2), we get $$a = 2$$ and $$d = 3$$ $$\therefore$$ AP is 2, 5, 8, 11, 14, 17 Now, Variance $$({\sigma ^2}) = {{\sum {x_i^2} } \over 6} - {\left( {\overline x } \right)^2}$$ $$ = {{{2^2} + {5^2} + {8^2} + {{11}^2} + {{14}^2} + {{17}^2}} \over 6} - {\left( {{{19} \over 2}} \right)^2}$$ $$ = {{105} \over 4}$$ $$\therefore$$ $$8{\sigma ^2} = 8 \times {{105} \over 4} = 210$$

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