Question

The sum of 6th and 10th term of AP is 92 and the sum of 8th and 13th ilterm is 112 find the a.p

Answer 1

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

Let assume that

↝ First term of an AP be a

and

↝ Common difference of an AP be d

According to statement,

↝ The sum of 6th and 10th term of AP is 92.

$\rm :\longmapsto\:a_6 + a_{10} = 92$

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\purple{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

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,

$\rm :\longmapsto\:a + (6 - 1)d + a + (10 - 1)d = 92$

$\rm :\longmapsto\:a + 5d + a + 9d = 92$

$\rm :\longmapsto\:2a + 14d = 92$

$\rm :\longmapsto\:2(a + 7d) = 92$

$\bf\implies \:\boxed{ \tt{ \: a + 7d = 46}} - - - - (1)$

According to statement again,

↝ The sum of 8th and 13th term is 112.

$\rm :\longmapsto\:a_8 + a_{13} = 112$

$\rm :\longmapsto\:a + (8 - 1)d + a + (13 - 1)d = 112$

$\rm :\longmapsto\:a + 7d + a + 12d = 112$

$\rm :\longmapsto\:2a + 19d= 112$

$\rm :\longmapsto\:2(46 - 7d) + 19d= 112$

$\rm :\longmapsto\:92- 14d + 19d= 112$

$\rm :\longmapsto\:5d= 112 - 92$

$\rm :\longmapsto\:5d= 20$

$\bf\implies \:\boxed{ \tt{ \: d \: = \: 4 \: \: }}$

↝ On substituting d = 4 in equation (1), we get

$\rm :\longmapsto\:a + 7 \times 4 = 46$

$\rm :\longmapsto\:a + 28= 46$

$\rm :\longmapsto\:a= 46 - 28$

$\bf\implies \:\boxed{ \tt{ \: a\: = \: 18 \: \: }}$

Hence, The required AP series is

$\bf\implies \:\boxed{ \tt{ \: 18,22,26,30, - - - - }}$

More to know :

↝ 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.