Question

Sum of pth,qth,rth terms of an arithmetic progression are a,b,c respectively,then prove that a/p(q-r)+b/q(r-p)+c/r(p-q)=0​

Answer 1

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

Let assume that First term of an AP is A and common difference is d.

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ 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.

So, According to statement,

It is given that,

$\rm :\longmapsto\:S_p = a$

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

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

Again,

$\rm :\longmapsto\:S_q = b$

$\rm :\longmapsto\:\dfrac{q}{2} \bigg(2 \:A\:+\:(q\:-\:1)\:d \bigg) = b$

$\rm :\longmapsto\:\dfrac{1}{2} \bigg(2 \:A\:+\:(q\:-\:1)\:d \bigg) = \dfrac{b}{q} - - - (2)$

Also,

$\rm :\longmapsto\:S_r = c$

$\rm :\longmapsto\:\dfrac{r}{2} \bigg(2 \:A\:+\:(r\:-\:1)\:d \bigg) = c$

$\rm :\longmapsto\:\dfrac{1}{2} \bigg(2 \:A\:+\:(r\:-\:1)\:d \bigg) = \dfrac{c}{r} - - - (3)$

Now, Consider,

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

On substituting the values from equation (1), (2) and (3),

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

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

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

$\rm \: = \: 0 + d(p - 1)(q - r) + d(q - 1)(r - p) + d(r - 1)(p - q)$

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

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

$\rm \: = \: \:d \times 0$

$\rm \: = \: \:0$

Hence, Proved

Additional Information :-

↝ 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.