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There are n AM's between 1,& 31 such that 7th mean: {( {n - 1}

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There are $n$ AM's between $1 &31$ such that $7 t h$ mean: $(n - 1)^{t h}$ mean $= 5 : 9,$ then find the value of $n .$

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There are

n

AM's between

1 &31

such that

7 t h

mean:

(n - 1)^{t h}

mean

= 5 : 9,

then find the value of

n .

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Text solutionVerified

Let the arithmetic mean inserted be $A 1, A 2, A 3, .... A m$ .

now , series is $1, A 1, A 2, A 3, .... A m, 31$

hence , $a = 1, T_{n} = 31, n = n + 2$

$T_{n} = a + (n - 1) d$

$31 = 1 + (n + 2 - 1) d$

$30 = (n + 1) d$

$d = \frac{30}{n + 1}$ _______(1)

given that $\frac{T _{7}}{T _{n} - 1} = \frac{5}{9}$

$\frac{a + ( 7 - 1 + 1 ) d}{a + ( n - 1 - 1 + 1 ) d} = \frac{5}{9}$

(here , $a + (7 - 1 + 1) d$ , +1 after 7-1 is because in the series the 1st A.M. is second in terms of series)

$\frac{a + 7 d}{a + ( n - 1 ) d} = \frac{5}{9}$

$9 a + 63 d = 5 a + 5 n d - 5 d$

$4 a + 68 d = 5 n d$

$4 a = d (5 n - 68)$

putting value of d frm (1)

by solving this you will get value of $n = 14$

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Question Text	There are $n$ AM's between $1 &31$ such that $7 t h$ mean: $(n - 1)^{t h}$ mean $= 5 : 9,$ then find the value of $n .$
Answer Type	Text solution:1
Upvotes	150

There are n AM's between 1&31 such that 7th mean: (n−1)th mean =5:9, then find the value of n.

Text solutionVerified

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There are $n$ AM's between $1 &31$ such that $7 t h$ mean: $(n - 1)^{t h}$ mean $= 5 : 9,$ then find the value of $n .$