Question

Between 1 and 31, m number have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7th and (m-1)th no. is 5:9 find the value of m.​

Answer 1

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

Given that,

• Between 1 and 31, m number have been inserted in such a way that the resulting sequence is an A.P.

Let assume that m numbers be

$\rm :\longmapsto\:A_1,A_2,A_3, - - - - - ,A_m$

so that

$\rm :\longmapsto\:1,A_1,A_2,A_3, - - - - - ,A_m,31 \: are \: in \: AP$

So, Here,

• First term = 1

• Number of terms, n = m + 2

• Last term, = 31

Let assume that common difference of an AP is 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,

$\rm :\longmapsto\:31 = 1 + (m + 2 - 1)d$

$\rm :\longmapsto\:31 - 1 = (m + 1)d$

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

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

Further, it is given that

$\red{\rm :\longmapsto\:\dfrac{A_7}{A_{m - 1}} = \dfrac{5}{9} }$

$\rm :\longmapsto\:\dfrac{a + 7d}{a + (m - 1)d} = \dfrac{5}{9}$

$\rm :\longmapsto\:9a + 63d = 5a + (5m - 5)d$

$\rm :\longmapsto\:9a - 5a = (5m - 5)d - 63d$

$\rm :\longmapsto\:4a = (5m - 5- 63)d$

$\rm :\longmapsto\:4a = (5m - 68)d$

$\rm :\longmapsto\:4 \times 1 = (5m - 68) \times \dfrac{30}{m + 1}$

$\rm :\longmapsto\:4 = (5m - 68) \times \dfrac{30}{m + 1}$

$\rm :\longmapsto\:4m + 4 = 150m - 2040$

$\rm :\longmapsto\:4m - 150m = - 2040 - 4$

$\rm :\longmapsto\:- 146m = - 2044$

$\rm \implies\:\boxed{ \tt{ \: m \: = \: 14 \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information :-

1. If a and b are two numbers, then Arithmetic Mean between a and b is given by

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: A \: = \: \frac{a \: + \: b}{2} \: }}}$

2. If a, b, c are in AP, then b is called Arithmetic mean between a and c.

3. If n Arithmetic mean are inserted between two numbers a and b, then sum of all Arithmetic mean is equals to n times the single Arithmetic mean between them.

$\rm :\longmapsto\:A_1 + A_2 + A_3 + - - - + A_n =n\bigg[ \dfrac{a + b}{2}\bigg]$