Math, asked by anveshabhoi123, 5 hours ago

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​

Answers

Answered by mathdude500
5

\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.
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