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.​

1 thought on “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.​”

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

   [tex]\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}[/tex]

   [tex]\large\underline{\sf{To\:show – }}[/tex]

   [tex] \: \: \: \: \: \: \bull \: \: \sf\: a(q – r) + b(r – p) + c(p – q) = 0[/tex]

   [tex]\large\underline{\sf{Solution-}}[/tex]

   Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

   ↝ nᵗʰ term of an arithmetic sequence is,

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

   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,

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

   ➣ It is given that pᵗʰ term is a.

   [tex]\rm :\longmapsto\:\begin{gathered}\bf{\boxed{ \sf \: a\:=\:A\:+\:(p\:-\:1)\:d}}–(i) \\ \end{gathered}[/tex]

   Aɢᴀɪɴ,

   ↝ qᵗʰ term is,

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

   ➣ It is given that qᵗʰ term is b.

   [tex]\rm :\longmapsto\:\begin{gathered}\bf{\boxed{ \sf \: b\:=\:A\:+\:(q\:-\:1)\:d}}–(ii) \\ \end{gathered}[/tex]

   Aɢᴀɪɴ,

   ↝ rᵗʰ term is,

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

   ➣ It is given that rᵗʰ term is c.

   [tex]\rm :\longmapsto\:\begin{gathered}\bf{\boxed{ \sf \: c\:=\:A\:+\:(r\:-\:1)\:d}}–(iii) \\ \end{gathered}[/tex]

   Now,

   ↝ Consider,

   [tex]\rm :\longmapsto\:\sf\: a(q – r) + b(r – p) + c(p – q)[/tex]

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

   [tex] \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) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [/tex]

   [tex] \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) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [/tex]

   [tex] \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) \: \: \: \: \: \: \: \: \: \: \: \: [/tex]

   [tex] \sf \: = 0 + d(pq – pr – r + p + qr – pq – r + p + rp – rq – p + q)[/tex]

   [tex] \sf \: = 0[/tex]

   [tex]{\boxed{\boxed{\bf{Hence, Proved}}}}[/tex]

   Additional Information :-

   Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

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

   [tex] \: \: \: \: \: \: \large\boxed{ \sf \: S_n \: = \: \dfrac{n}{2}\bigg(2a + (n – 1)d \bigg)} [/tex]

   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.

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