Question

The pth, qth&rth term of an A.P are a,b,c respectively prove that a(q-r)+b(r-p)+c(p-q)=0

Answer 1

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

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

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

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.

Let assume that

↝ First term of an AP = A

and

↝ Common difference of an AP = d

According to statement,

$\red{\rm :\longmapsto\:a_p = a}$

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

Also,

$\red{\rm :\longmapsto\:a_q = b}$

$\bf\implies \:\boxed{ \tt{ \: A + (q - 1)d = b}}$

and

$\red{\rm :\longmapsto\:a_r = c}$

$\bf\implies \:\boxed{ \tt{ \: A + (r - 1)d = c}}$

Now, Consider

$\rm :\longmapsto\:a(q-r)+b(r-p)+c(p-q)$

On substituting the values of a, b and c, we get

$\rm=[A + (p - 1)d](q - r) + [A + (q - 1)d](r - p) + [A + (r - 1)d](p - q)$

$\rm \: = \:A(q - r) + d(p - 1)(q - r) + \\ \rm \: A(r - p) + d(r - p)(q - 1) + \\ \rm \: A(p - q) + d(p - q)(r - 1)$

$\rm \: = \:A(q - r + r - p + p - q) + \\ \rm \: d[(p - 1)(q - r) + (q - 1)(r - p) + (r - 1)(p - q)]$

$\rm = 0 + d\bigg[pq - pr - q + r + qr - qp - r + p + rp - rq - p + q]$

$\rm \: = \:0$

Hence,

$\bf\implies \:\boxed{ \tt{ \: a(q-r)+b(r-p)+c(p-q) = 0}}$