Question

Solve this question and kindly explain the concept of insertion of n numbers between two given numbers.
Is there any formula to find the number is more than 1 numbers are inserted between two numbers in GP ?​

Answer 1

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

Given that,

m Arithmetic means are inserted between two numbers 1 and 31.

Let assume that m Arithmetic means be

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

So,

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

Now, here

First term of an AP, a = 1

Number of terms, n = m + 2

Last term, nth term = 31

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,

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

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

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

$\rm \implies\:d \: = \: \dfrac{30}{m + 1} - - - - (1)$

According to statement,

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

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

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

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

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

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

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

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

$\rm :\longmapsto\:2= (5m - 68) \times \dfrac{15}{m + 1}$

$\rm :\longmapsto\:2m + 2 = 75m - 1020$

$\rm :\longmapsto\:2m - 75m = - 2 - 1020$

$\rm :\longmapsto\: - 73m = - 1022$

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

• Hence, Option (d) is correct

Now,

Let assume that n geometric mean inserted between two numbers a and b be

$\rm :\longmapsto\:G_1,G_2,G_3, - - - - ,G_n$

So, that,

$\rm :\longmapsto\:a,G_1,G_2,G_3, - - - - ,G_n,b \: are \: in \: GP$

having common ratio

$\boxed{ \tt{ \: \: r \: = \: {\bigg[\dfrac{b}{a} \bigg]}^{\dfrac{1}{n + 1} } \: \: }}$

and

$\rm :\longmapsto\:G_1 = ar$

$\rm :\longmapsto\:G_2 = a {r}^{2}$

$\rm :\longmapsto\:G_3 = a {r}^{3}$

.

.

.

$\rm :\longmapsto\:G_n = a {r}^{n}$