Question

if the ratio of sum of n terms of two district A.P is as Sn and s'n is (n plus 5 by 2n minus 3 given below ) then find the ratio of 7th term ratio T7 / Ť7

$\binom{n + 5}{2n - 3}$
​

Answer 1

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

Let assume that

For the first AP series,

• First term of AP is a

• Common difference of an AP is d

• Number of terms is n

For the second AP series

• First term of AP is A

• Common difference of an AP is D

• Number of terms is n

According to statement

$\rm :\longmapsto\:\dfrac{S_n}{S_n'} = \dfrac{n + 5}{2n - 3}$

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ 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.

So, using this, we get

$\rm :\longmapsto\:\dfrac{\dfrac{n}{2} [2a + (n - 1)d]}{\dfrac{n}{2}[2A + (n - 1)D] } = \dfrac{n + 5}{2n - 3}$

$\rm :\longmapsto\:\dfrac{2a + (n - 1)d}{2A + (n - 1)D} = \dfrac{n + 5}{2n - 3}$

$\rm :\longmapsto\:\dfrac{a + (n - 1)\dfrac{d}{2} }{A + (n - 1)\dfrac{D}{2} } = \dfrac{n + 5}{2n - 3}$

Now,

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,

$\purple{\rm :\longmapsto\:Replace \: \dfrac{n - 1}{2} \: by \: 6, i.e. \: n \: by \: 13\: we \: get \: }$

$\rm :\longmapsto\:\dfrac{a + 6d }{A + 6D} = \dfrac{13 + 5}{26 - 3}$

$\rm :\longmapsto\:\dfrac{T_7}{T_7'} = \dfrac{18}{23}$

Answer 2

Answer:

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

Let assume that

For the first AP series,

First term of AP is a

Common difference of an AP is d

Number of terms is n

For the second AP series

First term of AP is A

Common difference of an AP is D

Number of terms is n

According to statement

$\rm :\longmapsto\:\dfrac{S_n}{S_n'} = \dfrac{n + 5}{2n - 3}$

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ 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.

So, using this, we get

$\rm :\longmapsto\:\dfrac{\dfrac{n}{2} [2a + (n - 1)d]}{\dfrac{n}{2}[2A + (n - 1)D] } = \dfrac{n + 5}{2n - 3}$

$\rm :\longmapsto\:\dfrac{2a + (n - 1)d}{2A + (n - 1)D} = \dfrac{n + 5}{2n - 3}$

$\rm :\longmapsto\:\dfrac{a + (n - 1)\dfrac{d}{2} }{A + (n - 1)\dfrac{D}{2} } = \dfrac{n + 5}{2n - 3}$

Now,

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,

$\purple{\rm :\longmapsto\:Replace \: \dfrac{n - 1}{2} \: by \: 6, i.e. \: n \: by \: 13\: we \: get \: }$

$\rm :\longmapsto\:\dfrac{a + 6d }{A + 6D} = \dfrac{13 + 5}{26 - 3}$

$\rm :\longmapsto\:\dfrac{T_7}{T_7'} = \dfrac{18}{23}$