Question

there are n A. M between 7 and 85 such that (n ._ 3)th mean : nth mean is as 11:24; find n

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

There are n A. M. between two numbers 7 and 85.

Let assume that

$\rm \: A_1, \: A_2, \: A_3, - - - A_n \: be \: n \: AM$

So, that

$\rm \:7, A_1, \: A_2, \: A_3, - - - A_n,85 \: are \: in \: AP$

So, we have

First term, a = 7

Number of terms = n + 2

nth term = 85

Let assume that Common difference of an AP be d

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

aₙ is the nᵗʰ term.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.

Tʜᴜs,

$\rm \: 85 = 7 + (n + 2 - 1)d$

$\rm \: 85 = 7 + (n + 1)d$

$\rm \: 85 - 7 = (n + 1)d$

$\rm \: 78 = (n + 1)d$

$\bf\implies \:d = \dfrac{78}{n + 1} - - - (1)$

According to statement,

$\rm \: \dfrac{A_{n - 3}}{A_n} \: = \: \dfrac{11}{24}$

$\rm \: \dfrac{a + (n - 3)d}{a + nd} = \dfrac{11}{24}$

$\rm \: 24a + 24d(n - 3) = 11a + 11nd$

$\rm \: 24a - 11a = 11nd - 24d(n - 3)$

$\rm \: 13a = (11n - 24n + 72)$

$\rm \: 13a = (72 - 13n)d$

On substituting the value of a and d, we get

$\rm \: 13 \times 7 = (72 - 13n) \times \dfrac{78}{n + 1}$

$\rm \: 7 = (72 - 13n) \times \dfrac{6}{n + 1}$

$\rm \: 7n + 7 = 432 - 78n$

$\rm \: 7n + 78n = 432 - 7$

$\rm \: 85n = 425$

$\bf\implies \:n \: = \: 5$

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ADDITIONAL INFORMATION

1. The sum of n AM's inserted between two numbers a and b is n times the single AM between a and b.

2. If A is the AM between two numbers a and b, then

$\rm \: A \: = \: \dfrac{a + b}{2}$