Question

The sum of first 11terms in an arithmetic sequence is 275.the sum of first 19 terms is 779.what is the 6th term

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.

According to statement,

↝ Sum of first 11 terms of an AP is 275.

$\rm :\longmapsto\:S_{11} = 275$

$\rm :\longmapsto\:\dfrac{\cancel{11}}{2} \bigg(2 \:a\:+\:(11\:-\:1)\:d \bigg) = \cancel{275} \: \: 25$

$\rm :\longmapsto\:\dfrac{1}{2} \bigg(2 \:a\:+\:10\:d \bigg) = \: 25$

$\rm :\longmapsto\:a + 5d = 25 - - - - (1)$

According to statement again,

↝ Sum of first 19 terms is 779

$\rm :\longmapsto\:S_{19} = 779$

$\rm :\longmapsto\:\dfrac{\cancel{19}}{2} \bigg(2 \:a\:+\:(19\:-\:1)\:d \bigg) = \cancel{779} \: \: 41$

$\rm :\longmapsto\:\dfrac{1}{2} \bigg(2 \:a\:+\:18\:d \bigg) = \: 41$

$\rm :\longmapsto\:a + 9d = 41 - - - - (2)$

Now,

↝ On Subtracting equation (1) from equation (2), we get

$\rm :\longmapsto\:4d = 16$

$\bf\implies \:d = 4$

↝ On substituting the value of d, in equation (1), we have

$\rm :\longmapsto\:a + 5(4) = 25$

$\rm :\longmapsto\:a + 20= 25$

$\rm :\longmapsto\:a = 25 - 20$

$\bf\implies \:a= 5$

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,

↝ 6ᵗʰ term is,

$\rm :\longmapsto\:a_6 = a + (6 - 1)d$

$\rm :\longmapsto\:a_6 = a + 5d$

On substituting the values of a and d, we get

$\rm :\longmapsto\:a_6 = 5 + 5(4)$

$\rm :\longmapsto\:a_6 = 5 +20$

$\bf :\longmapsto\:a_6 = 25$

Answer 2

Given that ,

• The sum of first 11 terms in an arithmetic sequence is 275 & the sum of first 19 terms is 779 .

Exigency To Find :

• The 6 th term of an A.P ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

¤ To Calculate the Sum total of an A.P is Given by

$\begin{gathered}\qquad \star\:\underline{\boxed {\pmb{\sf{ S_n \: =\: \dfrac{n}{2} \bigg\lgroup \sf{ 2a + ( n - 1 ) d }\bigg\rgroup}}}}\\\end{gathered}$

Here n is the n th Term of an A.P a is the first term of an A.P & d is common Difference of an A.P.

⠀⠀CASE I : The sum of first 11 terms in an arithmetic sequence is 275 .

$\begin{gathered} \qquad \dashrightarrow \sf S_n \: =\: \dfrac{n}{2} \bigg\lgroup \sf{ 2a + ( n - 1 ) d }\bigg\rgroup \\\\ \qquad \dashrightarrow \sf 275 \: =\: \dfrac{11}{2} \bigg\lgroup \sf{ 2a + ( 11 - 1 ) d }\bigg\rgroup \\\\ \qquad \dashrightarrow \sf 275 \: =\: \dfrac{11}{2} \bigg\lgroup \sf{ 2a + 10 d }\bigg\rgroup \\\\ \qquad \dashrightarrow \sf 275 \: =\: 11 \bigg\lgroup \sf{ a + 5 d }\bigg\rgroup \\\\ \qquad \dashrightarrow \sf a + 5d\: =\: 25 \:\qquad \bigg\lgroup \sf{ eq^n \:i }\bigg\rgroup \\\\\end{gathered}$

⠀⠀CASE II : The sum of first 19 terms in an arithmetic sequence is 779 .

$\begin{gathered} \qquad \dashrightarrow \sf S_n \: =\: \dfrac{n}{2} \bigg\lgroup \sf{ 2a + ( n - 1 ) d }\bigg\rgroup \\\\ \qquad \dashrightarrow \sf 779 \: =\: \dfrac{19}{2} \bigg\lgroup \sf{ 2a + ( 19 - 1 ) d }\bigg\rgroup \\\\ \qquad \dashrightarrow \sf 779 \: =\: \dfrac{19}{2} \bigg\lgroup \sf{ 2a + 18 d }\bigg\rgroup \\\\ \qquad \dashrightarrow \sf 779 \: =\: 19 \bigg\lgroup \sf{ a + 9 d }\bigg\rgroup \\\\ \qquad \dashrightarrow \sf a + 9d\: =\: 41 \:\qquad \bigg\lgroup \sf{ eq^n \:ii }\bigg\rgroup \\\\\end{gathered}$

We get ,

$\begin{gathered} \qquad \leadsto \sf a + 5d\: =\: 25 \:\qquad \bigg\lgroup \sf{ eq^n \:i }\bigg\rgroup \\\\\end{gathered}$

$\begin{gathered} \qquad \leadsto \sf a + 9d\: =\: 41 \:\qquad \bigg\lgroup \sf{ eq^n \:ii }\bigg\rgroup \\\\\end{gathered}$

$\begin{gathered}\dag \underline {\frak {By \: Subtracting\: eq^n\:i \: from \: eq^n \:ii\::}}\\\\\end{gathered}$

$\begin{gathered} \qquad \dashrightarrow \sf a + 9d\:- ( a + 5d ) =\: 41 \:- 25 \\\\\end{gathered}$

$\begin{gathered} \qquad \dashrightarrow \sf a + 9d\:- a - 5d =\: 16 \\\\\end{gathered}$

$\begin{gathered} \qquad \dashrightarrow \sf 4d =\: 16 \\\\\end{gathered}$

$\begin{gathered} \qquad \dashrightarrow \underline { \boxed { \pmb { \frak{ d \:( \: or \:Common \:Difference\:)\:=\: 4 \:}}}}\\\\\end{gathered}$

$\begin{gathered}\dag \underline {\frak{ By \: Substituting \: \:value \: of \: d \: eq^n\: i\::}}\\\\\end{gathered}$

$\begin{gathered} \qquad \leadsto \sf a + 5d\: =\: 25 \:\qquad \bigg\lgroup \sf{ eq^n \:i }\bigg\rgroup \\\\\end{gathered}$

$\begin{gathered} \qquad \dashrightarrow \sf a + 5d\: =\: 25 \: \\\\\end{gathered}$

$\begin{gathered} \qquad \dashrightarrow \sf a + 5(4)\: =\: 25 \: \\\\\end{gathered}$

$\begin{gathered} \qquad \dashrightarrow \sf a \: =\: 25 -20 \: \\\\\end{gathered}$

$\begin{gathered} \qquad \dashrightarrow \underline { \boxed { \pmb { \frak{ a \:( \: or \:First \:Term\:)\:=\: 5 \:}}}}\\\\\end{gathered}[/tex[</p><p></p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p></p><h3>⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀</h3><p></p><h3>⠀⠀⠀⠀⠀━ Finding 6 th term of an A.P :</h3><p></p><p>¤ <strong>To Calculate an n th terms of an A.P is Given by</strong> </p><p></p><p>[tex]\begin{gathered}\qquad \star\:\underline{\boxed {\pmb{\sf{ a_n \: =\: a + ( n - 1 ) d }}}}\\\\\end{gathered}$

⠀⠀⠀⠀⠀Here n is the n th Term of an A.P a is the first term of an A.P & d is common Difference of an A.P.

$\begin{gathered} \qquad \dashrightarrow \sf a_n \: =\: a + ( n - 1 ) d \\\\ \qquad \dashrightarrow \sf a_6 \: =\: 5 + ( 6 - 1 ) 4 \\\\ \qquad \dashrightarrow \sf a_6 \: =\: 5 + ( 5 ) 4 \\\\ \qquad \dashrightarrow \sf a_6 \: =\: 5 + 20 \\\\ \qquad \dashrightarrow \sf a_n \: =\: 25 \\\\ \qquad \dashrightarrow \underline { \boxed { \pmb { \frak{ a_6 \:( \: or \:Sixth \:Term\:)\:=\: 25 \:}}}}\\\\\end{gathered}$

$\begin{gathered}\qquad \therefore \:\underline {\sf Hence, \:6^{th}\:term\:of\:an\:A.P \:is\:\pmb{\bf 25\:}\:.}\\\\\end{gathered}$