Question

in an ap the sum of first 9 trems is 14 more than 5 timed the 8th trem 8th abd 2nd trems are in the ratio 11:2 find the sum of first 20 trems​

Answer 1

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

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.

And

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,

According to statement,

★ The sum of first 9 terms of an AP is 14 more than 5 times the 8th term.

$\rm :\longmapsto\:\dfrac{9}{2}\bigg(2a + (9 - 1)d\bigg) = 14 + 5 \bigg(a + (8 - 1)d \bigg)$

$\rm :\longmapsto\:\dfrac{9}{2}\bigg(2a + 8d\bigg) = 14 + 5 \bigg(a + 7d \bigg)$

$\rm :\longmapsto\:9(a + 4d) = 14 + 5a + 35d$

$\rm :\longmapsto\:9a + 36d = 14 + 5a + 35d$

$\rm :\implies\:d = 14 - 4a - - - (1)$

According to second condition,

★ Ratio of 8th term and 2nd term is 11 : 2

$\rm :\longmapsto\:\dfrac{a + (8 - 1)d}{a + (2 - 1)d} = \dfrac{11}{2}$

$\rm :\longmapsto\:\dfrac{a + 7d}{a + d} = \dfrac{11}{2}$

$\rm :\longmapsto\:11a + 11d = 2a + 14d$

$\rm :\longmapsto\:11a - 2a = 14d - 11d$

$\rm :\longmapsto\:9a = 3d$

$\rm :\longmapsto\:3a = d$

$\rm :\longmapsto\:3a = 14 - 4a \: \: \: \: \: \: \{ \: using \: (1) \: \}$

$\rm :\longmapsto\:3a + 4a = 14$

$\rm :\longmapsto\:7a = 14$

$\bf\implies \:a = 2$

On substituting the value of a, in equation (1), we get

$\rm :\implies\:d = 14 - 4(2)$

$\rm :\implies\:d = 14 - 8$

$\bf\implies \:d = 6$

So,

We have now,

• First term of an AP, a = 2

• Common difference of an AP, d = 6

• Number of terms,n = 20

We know,

${{{{{{\rm :\longmapsto\:S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}}$

${{{{{{\rm :\longmapsto\:S_{20}\:=\dfrac{20}{2} \bigg(2 \:(2)\:+\:(20\:-\:1)\:6 \bigg)}}}}}}$

${{{{{{\rm :\longmapsto\:S_{20}\:=10 \bigg(4\:+\:(19)\:6 \bigg)}}}}}}$

${{{{{{\rm :\longmapsto\:S_{20}\:=10 \bigg(4\:+ \: 76\bigg)}}}}}}$

$\rm :\longmapsto\:S_{20} = 800$