Question

there are n arithmetic means between 5 and 35. If the second mean:the last mean=1:4,find n​

Answer 1

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5 and 35. If the second mean:the last mean=1:4,find n

Answer 2

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

Let assume that

$\rm :\longmapsto\:A_1,A_2,A_3, - - - - - - ,A_n \: be \: n \: arithmetic \: means$

between 5 and 35.

So,

$\rm :\longmapsto\:5,A_1,A_2,A_3, - - - -- ,A_n ,35 \: are \: in \: AP$

So, we have

↝ First term of an AP, a = 5

↝ Number of terms = n + 2

↝ Last term of an AP = 35

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,

On substituting the values, we get

$\rm :\longmapsto\:35 = 5 + (n + 2 - 1)d$

$\rm :\longmapsto\:35 - 5 = (n + 1)d$

$\rm :\longmapsto\:30 = (n + 1)d$

$\rm \implies\:\boxed{ \tt{ \: d \: = \: \frac{30}{n + 1} \: \: }}$

Now, According to statement, it is given that

$\rm :\longmapsto\:\dfrac{A_2}{A_n} = \dfrac{1}{4}$

$\rm :\longmapsto\:\dfrac{a + 2d}{a + nd} = \dfrac{1}{4}$

$\rm :\longmapsto\:4a + 8d = a + nd$

$\rm :\longmapsto\:4a - a = nd - 8d$

$\rm :\longmapsto\:3a = (n - 8)d$

On substituting the values of a and d, we get

$\rm :\longmapsto\:3 \times 5 = (n - 8) \times \dfrac{30}{n + 1}$

$\rm :\longmapsto\:15 = (n - 8) \times \dfrac{30}{n + 1}$

$\rm :\longmapsto\:1 = (n - 8) \times \dfrac{2}{n + 1}$

$\rm :\longmapsto\:n + 1 = 2n - 16$

$\rm :\longmapsto\:n - 2n = - 1 - 16$

$\rm :\longmapsto\:- n = - 17$

$\rm \implies\:\boxed{ \tt{ \: \: n \: \: = \: \: 17 \: \: }}$

Additional Information :-

Let a and b are two positive real numbers, then

$\rm :\longmapsto\:\boxed{ \tt{ \: AM \: = \: \dfrac{a + b}{2} \: }}$

$\rm :\longmapsto\:\boxed{ \tt{ \: GM \: = \: \sqrt{ab} \: }}$

$\rm :\longmapsto\:\boxed{ \tt{ \: HM \: = \: \frac{2ab}{a \: + \: b} \: }}$

$\rm :\longmapsto\:\boxed{ \tt{ \: AM \: \geqslant \: GM \: \geqslant \: HM \: \: }}$