Question

if the 3rd term of a GP is 27 and the 12th term is 315, find the GP​

Answer 1

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

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an geometric sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a {r}^{n - 1} }}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.

• a is the first term of the sequence.

• n is the no. of terms.

• r is the common ratio.

Tʜᴜs,

↝ 3ᵗʰ term is,

$\rm :\longmapsto\:a_3\:=\:a {r}^{3 - 1}$

$\rm :\longmapsto\:a_3\:=\:a {r}^{2}$

➣ It is given that 3ᵗʰ term is 27.

$\bf\implies \: {ar}^{2} = 27 - - - (1)$

Aɢᴀɪɴ,

↝ 12ᵗʰ term is,

$\rm :\longmapsto\:a_{12}\:=\:a {r}^{12 - 1}$

$\rm :\longmapsto\:a_{12}\:=\:a {r}^{11}$

➣ It is given that m12ᵗʰ term is 315.

$\bf\implies \: {ar}^{11} = 315 - - - (2)$

Now,

On dividing equation (1) by equation (2), we get

$\rm :\longmapsto\:\dfrac{ {ar}^{11} }{ {ar}^{2} } = \dfrac{315}{27}$

$\rm :\longmapsto\: {r}^{9} = \dfrac{35}{3}$

$\bf\implies \:r = {\bigg(\dfrac{35}{3} \bigg) }^{\dfrac{1}{9} }$

On substituting the value of r in equation (1), we get

$\rm :\longmapsto\:a{\bigg(\dfrac{35}{3} \bigg) }^{\dfrac{2}{9} } = 27$

$\rm :\longmapsto\:a = 27 \times {\bigg(\dfrac{35}{3} \bigg) }^{\dfrac{ - 2}{9} }$

Hence,

$\rm :\longmapsto\:a_1\: = 27 {\bigg(\dfrac{35}{3} \bigg) }^{\dfrac{ - 2}{9} }$

$\rm :\longmapsto\:a_2\: =ar = 27 {\bigg(\dfrac{35}{3} \bigg) }^{\dfrac{ - 2}{9}} \times {\bigg(\dfrac{35}{3} \bigg) }^{\dfrac{1}{9}} = 27 {\bigg(\dfrac{35}{3} \bigg) }^{\dfrac{ - 1}{9}}$

$\rm :\longmapsto\:a_3\: = 27$

Therefore,

The required GP series is

$\rm :\longmapsto\:27{\bigg(\dfrac{35}{3} \bigg) }^{\dfrac{ - 2}{9} }, \: 27{\bigg(\dfrac{35}{3} \bigg) }^{\dfrac{ - 1}{9} },27, - - - -$

Additional Information :-

Sum of n terms of Geometric Progression

$\begin{gathered}\begin{gathered}\bf\: S_n = \begin{cases} &\sf{a\bigg(\dfrac{ {r}^{n} - 1}{r - 1} \bigg)} \: if \: r \ne \: 1\\ &\sf{na \: \: if \: r = 1} \end{cases}\end{gathered}\end{gathered}$