मराठी

In N A.M.'S Are Introduced Between 3 and 17 Such that the Ratio of the Last Mean to the First Mean is 3 : 1, Then the Value of N is - Mathematics

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प्रश्न

In n A.M.'s are introduced between 3 and 17 such that the ratio of the last mean to the first mean is 3 : 1, then the value of n is

पर्याय

  • 6

  • 8

  • 4

  •  none of these.

MCQ
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उत्तर

6

Let 

\[A_1 , A_2 , A_3 , A_4 . . . . A_n\] be the n arithmetic means between 3 and 17.
Let d be the common difference of the A.P. 3,

\[A_1 , A_2 , A_3 , A_4 , . . . . A_n\] and 17.
Then, we have:

d = \[\frac{17 - 3}{n + 1}\] = \[\frac{14}{n + 1}\]

Now, 

\[A_1\] = 3 + d = 3 + \[\frac{14}{n + 1}\] = \[\frac{3n + 17}{n + 1}\]

And, 

\[A_n = 3 + nd = 3 + n\left( \frac{14}{n + 1} \right) = \frac{17n + 3}{n + 1}\]

\[\therefore \frac{A_n}{A_1} = \frac{3}{1}\]

\[ \Rightarrow \frac{\left( \frac{17n + 3}{n + 1} \right)}{\left( \frac{3n + 17}{n + 1} \right)} = \frac{3}{1}\]

\[ \Rightarrow \frac{17n + 3}{3n + 17} = \frac{3}{1}\]

\[ \Rightarrow 17n + 3 = 9n + 51\]

\[ \Rightarrow 8n = 48\]

\[ \Rightarrow n = 6\]

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पाठ 19: Arithmetic Progression - Exercise 19.9 [पृष्ठ ५१]

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आरडी शर्मा Mathematics [English] Class 11
पाठ 19 Arithmetic Progression
Exercise 19.9 | Q 5 | पृष्ठ ५१

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