Question

if seven times of seven th term is equal to eleven times of eleven th term show that eighteen th term is zero​

Answer 1

Step-by-step explanation:

According to question,

7×7t = 11×11t

This is only possible when both the terms are equal to 0.

Therefore,

7×7t = 0

=> 7t = 0/7 = 0

=> t = 0/7 = 0

=> t = 0

Therefore,

18t = 18×0 = 0

Hence, Proved.

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Answer 2

Appropriate Question :-

if seven times of seventh term of an AP is equal to eleven times of eleventh term, show that eighteenth term is zero.

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

Let assume that first term and common difference of an AP is a and d respectively.

According to statement, it is given that 7 times the $7^{th}$ term of an AP is equals to 11 times the $11^{th}$ term of an AP.

So,

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

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic progression 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 progression.

• n is the no. of terms.

• d is the common difference.

Tʜᴜs,

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

$\rm \: 7(a + 6d) = 11(a + 10d)$

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

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

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

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

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

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

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

Hence, Proved

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

Additional Information :-

↝ Sum of n  terms of an arithmetic progression 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 progression.

• n is the no. of terms.

• d is the common difference.