Math, asked by nitishmallik12, 1 month ago

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

Answers

Answered by mathdude500
5

\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} }

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