Question

1. The pth, qth, rth terms of AP are a, b,c respectively. Show that: a(q + r) + b(r – p) + c(p – 9) = 0.​

Answer 1

Correct Statement

The pth, qth, rth terms of AP are a, b,c respectively. Show that: a(q - r) + b(r – p) + c(p – q) = 0.

Answer

$\begin{gathered}\begin{gathered}\bf \:Given-\begin{cases} &\sf{a_p = a} \\ &\sf{a_q = b}\\ &\sf{a_r = c} \end{cases}\end{gathered}\end{gathered}$

$\large\underline{\sf{To\:show - }}$

$\: \: \: \: \: \: \bull \: \: \sf\: a(q - r) + b(r - p) + c(p - q) = 0$

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

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\:\:{\underline{{\boxed{\bf{{a_n\:=\:A\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

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,

↝ pᵗʰ term is,

$\rm :\longmapsto\:\begin{gathered}\bf{a_p\:=\:A\:+\:(p\:-\:1)\:d} \\ \end{gathered}$

➣ It is given that pᵗʰ term is a.

$\rm :\longmapsto\:\begin{gathered}\bf{\boxed{ \sf \: a\:=\:A\:+\:(p\:-\:1)\:d}}--(i) \\ \end{gathered}$

Aɢᴀɪɴ,

↝ qᵗʰ term is,

$\rm :\longmapsto\:\begin{gathered}\bf{a_q\:=\:A\:+\:(q\:-\:1)\:d} \\ \end{gathered}$

➣ It is given that qᵗʰ term is b.

$\rm :\longmapsto\:\begin{gathered}\bf{\boxed{ \sf \: b\:=\:A\:+\:(q\:-\:1)\:d}}--(ii) \\ \end{gathered}$

Aɢᴀɪɴ,

↝ rᵗʰ term is,

$\rm :\longmapsto\:\begin{gathered}\bf{a_r\:=\:A\:+\:(r\:-\:1)\:d} \\ \end{gathered}$

➣ It is given that rᵗʰ term is c.

$\rm :\longmapsto\:\begin{gathered}\bf{\boxed{ \sf \: c\:=\:A\:+\:(r\:-\:1)\:d}}--(iii) \\ \end{gathered}$

Now,

↝ Consider,

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

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

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

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

$\sf \: = A\bigg( \cancel{p} - \cancel{q} + \cancel{q} - \cancel{r} + \cancel{r} - \cancel{p} \bigg) + d(pq - pr - q + r) + \\ \sf \: d(qr - pq - r + p) + d(rp - rq - p + q) \: \: \: \: \: \: \: \: \: \: \: \:$

$\sf \: = 0 + d(pq - pr - r + p + qr - pq - r + p + rp - rq - p + q)$

$\sf \: = 0$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ Sum of first n terms of an arithmetic sequence is,

$\: \: \: \: \: \: \large\boxed{ \sf \: S_n \: = \: \dfrac{n}{2}\bigg(2a + (n - 1)d \bigg)}$

Wʜᴇʀᴇ,

• Sₙ is the sum of first n terms.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.