Question

In A.P Tp = 1 by q and Tq = 1 by p then a-d =

Answer 1

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

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,

Given that,

$\red{\rm :\longmapsto\:T_p = \dfrac{1}{q}}$

$\red{\rm :\longmapsto\:a + (p - 1)d = \dfrac{1}{q} - - - (1)}$

Also, Given that

$\red{\rm :\longmapsto\:T_q = \dfrac{1}{p}}$

$\red{\rm :\longmapsto\:a + (q - 1)d = \dfrac{1}{p} - - - (2)}$

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

$\rm :\longmapsto\:(p - 1)d - (q - 1)d = \dfrac{1}{q} - \dfrac{1}{p}$

$\rm :\longmapsto\:pd - d - qd + d = \dfrac{p - q}{pq}$

$\rm :\longmapsto\:d(p - q) = \dfrac{p - q}{pq}$

$\bf\implies \:d = \dfrac{1}{pq} - - - (3)$

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

$\rm :\longmapsto\:a + (p - 1)\dfrac{1}{pq} = \dfrac{1}{q}$

$\rm :\longmapsto\:a + \dfrac{1}{q} - \dfrac{1}{pq} = \dfrac{1}{q}$

$\rm :\longmapsto\:a - \dfrac{1}{pq} = 0$

$\bf\implies \:\:a = \dfrac{1}{pq}$

Hence,

$\bf\implies \:\:a - d = \dfrac{1}{pq} - \dfrac{1}{pq} = 0$

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.