Question

if 7times of the 7th term of AP is equal to the 11times kf the 11th term, find the 18th term​

Answer 1

Answer:

here's explanation in attachment

Answer 2

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

Given that,

• 7 times of the 7th term of AP is equal to the 11 times the 11th term.

i.e.

$\rm \: 7a_7 = 11a_{11}$

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,

$\rm \: 7[a + (7 - 1)d] = 11[a + (11 - 1)d]$

$\rm \: 7[a + 6d] = 11[a + 10d]$

$\rm \: 7a + 42d = 11a + 110d$

can be rewritten as

$\rm \: 11a + 110d - 7a - 42d = 0$

$\rm \: 4a + 68d= 0$

$\rm \: 4(a + 17d)= 0$

$\rm \: a + 17d= 0$

$\rm \: a + (18 - 1)d= 0$

$\rm\implies \:a_{18} = 0$

$\rule{190pt}{2pt}$

Short Cut Trick :-

If m times the mth term of an AP is equal to n times the nth term of an AP, then (m + n)th term of an AP = 0.

$\rule{190pt}{2pt}$

Additional Information :-

↝ 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.