Question

34. Between 1 and 31 m numbers have been inserted in such a way that the
resulting sequence is an AP and the ratio of 7th and (m-1) "" numbers is 5:9.
Find the value of m.​

Answer 1

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

$\rm :\longmapsto\:Let \: A_1,A_2,A_3, - - - ,A_m \: be \: m \: numbers$

$\rm \: inserted \: between \: 1 \: and \: 31 \: so \: that \: resultant \: series \: is \: AP$

So, it implies,

$\rm :\longmapsto\:3,A_1,A_2,A_3, - - - ,A_m,31 \: are \: in \:AP$

We know that,

↝ nᵗʰ term of an arithmetic sequence is,

$\rm :\longmapsto\:\begin{gathered}\:\:{\underline{{\boxed{\bf{{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

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.

ʜᴇʀᴇ,

• aₙ = 31

• a = 1

• n = m + 2

• Let d is the common difference.

Thus,

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

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

$\bf\implies \:d = \dfrac{30}{m + 1}$

According to statement,

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

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

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

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

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

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

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

$\bf\implies \:m \: = \: 14$