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Question

If $$a_1, a_2, a_3, a_4, a_5 \, and \, a_6$$ are consecutive odd natural numbers such that $$a_6 > a_1, \, a_1 + a_2 + a_3 + a_4 + a_5 + a_6 = p^3 \, and \, a_5 = 39$$, then the value of $$(a_3 + a_4 - 12p)$$ is equal to


A
1
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B
0
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C
2
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D
4
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Solution

The correct option is A 1
Given : $$a_{1},a_{2},a_{3},a_{4},a_{5} $$ and $$a_{6}$$ are consecutive odd natural numbers.
$$a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}=P^{3}$$ 
$$a_{r}=39(a_{4}=a_{5}-2,a_{3}=a_{4}-2,a_{2}=a_{3}-2,a_{1}=a_{2}-2,a_{6}+2)$$ 
$$\therefore a_{1}=31$$ $$a_{2}=33$$ $$a_{3}=35$$ $$a_{4}=37$$ $$a_{6}=39$$ 
$$31+33+35+37+39+41=216=6^{3}$$ 
      $$P=6$$ 
$$a_{3}+a_{4}-12P=35+37-12\times 6$$
                          $$=72-72$$

$$a_{3}+a_{4}-12P=0$$

Mathematics

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