Question

In a G.P, the 5th term exceeds the 4th term by 24, and the 4th term exceeds the 3rd term by 8, find the common ratio and the first term.​

Answer 1

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

Let assume that

➢ First term of a GP series is a

➢ Common ratio of GP series is r

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an geometric sequence is,

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

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,

According to statement,

$\rm :\longmapsto\:a_5 - a_4 = 24$

$\rm :\longmapsto\: {ar}^{5 - 1} - {ar}^{4 - 1} = 24$

$\rm :\longmapsto\: {ar}^{4} - {ar}^{3} = 24$

$\rm :\longmapsto\: {ar}^{3}(r - 1) = 24 - - - - (1)$

Again, According to statement

$\rm :\longmapsto\:a_4 - a_3 = 8$

$\rm :\longmapsto\:{ar}^{3} -{ar}^{2} = 8$

$\rm :\longmapsto\:{ar}^{2}(r - 1) = 8 - - - - (2)$

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

$\rm :\longmapsto\:\dfrac{{ar}^{3}(r - 1)}{{ar}^{2}(r - 1)} = 3$

$\rm\implies \:\boxed{\tt{ \: \: r = 3 \: \: }}$

On substituting the value of r in equation (2), we get

$\rm :\longmapsto\: 9a(3 - 1) = 8$

$\rm :\longmapsto\: 18a= 8$

$\rm\implies \:\boxed{\tt{ \: \: a \: = \: \frac{4}{9} \: \: }}$

Hence

$\begin{gathered}\begin{gathered}\bf\:\rm\implies \:\begin{cases} &\sf{r \: = \: 3 \: } \\ \\ &\sf{a \: = \: \dfrac{4}{9} \: } \end{cases}\end{gathered}\end{gathered}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

MORE TO KNOW

↝ Sum of n  terms of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{a \: ( {r}^{n} \: - \: 1)}{r \: - \: 1}\bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

Sₙ is the sum of n terms of AP.

a is the first term of the sequence.

n is the no. of terms.

r is the common ratio.

Answer 2

Answer:

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

Let assume that

➢ First term of a GP series is a

➢ Common ratio of GP series is r

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an geometric sequence is,

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

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,

According to statement,

$\rm :\longmapsto\:a_5 - a_4 = 24$

$\rm :\longmapsto\: {ar}^{5 - 1} - {ar}^{4 - 1} = 24$

$\rm :\longmapsto\: {ar}^{4} - {ar}^{3} = 24$

$\rm :\longmapsto\: {ar}^{3}(r - 1) = 24 - - - - (1)$

Again, According to statement

$\rm :\longmapsto\:a_4 - a_3 = 8$

$\rm :\longmapsto\:{ar}^{3} -{ar}^{2} = 8$

$\rm :\longmapsto\:{ar}^{2}(r - 1) = 8 - - - - (2)$

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

$\rm :\longmapsto\:\dfrac{{ar}^{3}(r - 1)}{{ar}^{2}(r - 1)} = 3$

$\rm\implies \:\boxed{\tt{ \: \: r = 3 \: \: }}$

On substituting the value of r in equation (2), we get

$\rm :\longmapsto\: 9a(3 - 1) = 8$

$\rm :\longmapsto\: 18a= 8$

$\rm\implies \:\boxed{\tt{ \: \: a \: = \: \frac{4}{9} \: \: }}$

Hence

$\begin{gathered}\begin{gathered}\bf\:\rm\implies \:\begin{cases} &\sf{r \: = \: 3 \: } \\ \\ &\sf{a \: = \: \dfrac{4}{9} \: } \end{cases}\end{gathered}\end{gathered}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬