Question

if pth qth and rth term of an ap is a b and c then show that a(q-r)+ b(r-p) +c(​

Answer 1

Correct Statement:-

• if pth qth and rth term of an ap is a b and c then show that a(q - r)+ b(r - p) +c(p - q) = 0

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Gɪᴠᴇɴ :

• The pᵗʰ term of an arithmetic sequence is a.
• The qᵗʰ term of an arithmetic sequence is b.
• The rᵗʰ term of an arithmetic sequence is c.

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To show :-

• a(q - r)+ b(r - p) +c(p - q) = 0

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Formula used :-

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.

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Solution :-

$\begin{gathered}\begin{gathered}\bf Let = \begin{cases} &\sf{first \: term \: of \: ap \: is \: x} \\ &\sf{common \: difference \: of \: ap \: = y} \end{cases}\end{gathered}\end{gathered}$

$\large\underline{\bold{❥︎Step :- 1 }}$

☆ Now, The pᵗʰ term of an arithmetic sequence is a.

$\tt \: \implies \: a \: = x + (p - 1)y - - - - (1)$

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$\large\underline{\bold{❥︎Step :- 2 }}$

☆ Again, The qᵗʰ term of an arithmetic sequence is b.

$\tt \: \implies \: b = x + (q - 1)y - - - (2)$

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$\large\underline{\bold{❥︎Step :- 3 }}$

Also, The rᵗʰ term of an arithmetic sequence is c.

$\tt \: \implies \: c = x + (r - 1)d - - - - (4)$

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$\large\underline{\bold{❥︎Step :- 4 }}$

$\tt \: Consider \: a(q - r)+ b(r - p) +c(p - q)$

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

$\tt \: \bigg( x + (p - 1)y\bigg)(q - r) + \bigg( x + (q - 1)y\bigg)(r - p) + \bigg( x + (r - 1)y\bigg)(p - q)$

$\tt \: = x(q - r) + (p \: - 1)y(q - r) + x(r - p) \tt \: + \\ \tt \: y(q - 1)(r - p) + x(p - q) + y(p - q)(r - 1)$

$\tt \: = x(q - r + r - p+ p - q) + y(pq - q - rp + r + \\ \tt \: qr - r - pr + r + pr - p - qr + q)$

$\tt \: = x(0) + y(0)$

$\tt \: = 0$

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

$\large{\boxed{\boxed{\tt{Hence, Proved}}}}$