Question

How many terms are there in a geometric sequence in which the first and last terms are 16 and 1/64 respectively and r=1/2​

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,

According to statement, In an GP, we are given that

$\red{\rm :\longmapsto\:a_1 = 16}$

$\red{\rm :\longmapsto\:a_n = \dfrac{1}{64} }$

$\red{\rm :\longmapsto\:r = \dfrac{1}{2} }$

Now, we have

$\rm :\longmapsto\:{a_n\:=\: {a \: r}^{n \: - \: 1}}$

On substituting the values, we get

$\rm :\longmapsto\:\dfrac{1}{64} = 16 {\bigg(\dfrac{1}{2} \bigg) }^{n - 1}$

$\rm :\longmapsto\:\dfrac{1}{64 \times 16} = {\bigg(\dfrac{1}{2} \bigg) }^{n - 1}$

$\rm :\longmapsto\:\dfrac{1}{ {2}^{6} \times {2}^{4} } = {\bigg(\dfrac{1}{2} \bigg) }^{n - 1}$

$\rm :\longmapsto\:\dfrac{1}{ {2}^{6 + 4} } = {\bigg(\dfrac{1}{2} \bigg) }^{n - 1}$

$\rm :\longmapsto\:\dfrac{1}{ {2}^{10} } = {\bigg(\dfrac{1}{2} \bigg) }^{n - 1}$

$\rm :\longmapsto\: {\bigg(\dfrac{1}{2} \bigg) }^{10 } = {\bigg(\dfrac{1}{2} \bigg) }^{n - 1}$

So, on comparing, we get

$\rm :\longmapsto\:n - 1 = 10$

$\bf :\longmapsto\:n = 11$

Additional Information :-

↝ Sum of n  terms of an geometric sequence is,

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

Wʜᴇʀᴇ,

Sₙ is the sum of n terms of GP.

a is the first term of the sequence.

n is the no. of terms.

r is the common ratio.