Question

The sum of n terms of two arithmetic progressions are in the ratio (3n+8): (7n + 12). Find the ratio of their 10th terms.​

Answer 1

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

Given that,

• The sum of n terms of two arithmetic progressions are in the ratio (3n+8): (7n + 12).

Let assume that

• First term of first AP is a

• Common difference of first AP is d

• First term of second AP is A

• Common difference of second AP is D

So, According to statement,

$\red{\rm :\longmapsto\:\dfrac{S_n}{S_n'} = \dfrac{3n + 8}{7n + 12} }$

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,

$\rm :\longmapsto\:\dfrac{\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}{\dfrac{n}{2} \bigg(2 \:A\:+\:(n\:-\:1)\:D \bigg)} = \dfrac{3n + 8}{7n + 12}$

$\rm :\longmapsto\:\dfrac{ 2 \:a\:+\:(n\:-\:1)\:d}{2 \:A\:+\:(n\:-\:1)\:D} = \dfrac{3n + 8}{7n + 12} - - - (1)$

Now, we have to find the ratio of 10th term of two series.

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,

$\red{\rm :\longmapsto\:\dfrac{a_{10}}{a_{10}} = \dfrac{a + 9d}{A + 9D} = \dfrac{2a + 18d}{2A + 18D} - - - (2)}$

Now, From equation (1), we have

$\rm :\longmapsto\:\dfrac{ 2 \:a\:+\:(n\:-\:1)\:d}{2 \:A\:+\:(n\:-\:1)\:D} = \dfrac{3n + 8}{7n + 12}$

Put n = 19, we get

$\rm :\longmapsto\:\dfrac{ 2 \:a\:+\:(19\:-\:1)\:d}{2 \:A\:+\:(19\:-\:1)\:D} = \dfrac{3(19) + 8}{7(19) + 12}$

$\rm :\longmapsto\:\dfrac{ 2 \:a\:+ \: 18\:d}{2 \:A\:+\:18\:D} = \dfrac{57 + 8}{133 + 12}$

$\rm :\longmapsto\:\dfrac{ 2 \:a\:+ \: 18\:d}{2 \:A\:+\:18\:D} = \dfrac{65}{145}$

$\rm :\longmapsto\:\dfrac{ 2 \:a\:+ \: 18\:d}{2 \:A\:+\:18\:D} = \dfrac{13}{29} - - - (3)$

From equation (2) and (3), we get

$\red{\rm :\longmapsto\:\dfrac{a_{10}}{a_{10}} = \dfrac{a + 9d}{A + 9D} = \dfrac{2a + 18d}{2A + 18D} = \dfrac{13}{29} }$