Question

In an arithemetic sequence, ratio of its 3rd and 6th term is 3:4 a) find the ratio of its 6th and 10th term.

Answer 1

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

Given that, In an AP sequence, the ratio of 3rd term and 6th term is 3 : 4.

Let assume that first term and common difference of an AP sequence be a and d respectively.

So,

$\rm \: \dfrac{a_3}{a_6} = \dfrac{3}{4}$

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,

$\rm \: \dfrac{a + (3 - 1)d}{a + (6 - 1)d} = \dfrac{3}{4}$

$\rm \: \dfrac{a + 2d}{a + 5d} = \dfrac{3}{4}$

$\rm \: 4a + 8d = 3a + 15d$

$\rm \: 4a - 3a = 15d - 8d$

$\rm\implies \:a \: = \: 7d$

Now, Consider

$\rm \: \dfrac{a_6}{a_{10}}$

$\rm \: = \: \dfrac{a + 5d}{a + 9d}$

On substituting the value of a, we get

$\rm \: = \: \dfrac{7d + 5d}{7d + 9d}$

$\rm \: = \: \dfrac{12d}{16d}$

$\rm \: = \: \dfrac{3}{4}$

$\rm\implies \:a_6 : a_{10} \: = \: 3 : 4$

$\rule{190pt}{2pt}$

Additional Information :-

↝ Sum of n  terms of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

Sₙ is the sum of n terms of AP.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.