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Question: If the sum of first p terms, first q terms and first r terms of an A.P be x, y and z respectively, t...

If the sum of first p terms, first q terms and first r terms of an A.P be x, y and z respectively, then xp(q−r)+yq(r−p)+zr(p−q)\dfrac{x}{p}(q-r)+\dfrac{y}{q}(r-p)+\dfrac{z}{r}(p-q) is

Explanation

Solution

Use the sum of n terms formula in A.P which is given by Sn=n2[2a+(n−1)d]{{S}_{n}}=\dfrac{n}{2}[2a+(n-1)d] ; where aa is the first term of A.P and d is the common difference. Try to find the value of xp,yq,zr\dfrac{x}{p},\dfrac{y}{q},\dfrac{z}{r} using the sum of n terms formula. Substitute the values so obtained in xp(q−r)+yq(r−p)+zr(p−q)\dfrac{x}{p}(q-r)+\dfrac{y}{q}(r-p)+\dfrac{z}{r}(p-q) .

Complete step by step answer:
We are given in the question that sum of first p terms Sp{{S}_{p}} , first q terms Sq{{S}_{q}} and first r terms Sr{{S}_{r}} be x, y and z respectively, so
\left. \begin{aligned} & {{S}_{p}}=x \\\ & {{S}_{q}}=y \\\ & {{S}_{r}}=z \\\ \end{aligned} \right\\}......(1)
We know that sum of n terms in A.P is given by Sn=n2[2a+(n−1)d]{{S}_{n}}=\dfrac{n}{2}[2a+(n-1)d] ; where aa is the first term of A.P and d is the common difference. Using this formula of sum of n term and equation (1) can be written as
Sp=p2[2a+(p−1)d]=x......(2){{S}_{p}}=\dfrac{p}{2}[2a+(p-1)d]=x......(2)
Sq=q2[2a+(q−1)d]=y......(3){{S}_{q}}=\dfrac{q}{2}[2a+(q-1)d]=y......(3)
Sr=r2[2a+(r−1)d]=z......(4){{S}_{r}}=\dfrac{r}{2}[2a+(r-1)d]=z......(4)
From equation (2) we will divide p throughout the equation and divide 2 inside the bracket in left hand side of equation we get,
xp=a+(p−1)d2......(5)\dfrac{x}{p}=a+(p-1)\dfrac{d}{2}......(5)
Similarly from equation (3) we have,
yq=a+(q−1)d2......(6)\dfrac{y}{q}=a+(q-1)\dfrac{d}{2}......(6)
Similarly from equation (4) we have,
zr=a+(r−1)d2.......(7)\dfrac{z}{r}=a+(r-1)\dfrac{d}{2}.......(7)
Now we will substitute the values of equation (5), (6) and (7) in xp(q−r)+yq(r−p)+zr(p−q)\dfrac{x}{p}(q-r)+\dfrac{y}{q}(r-p)+\dfrac{z}{r}(p-q) we get,
xp(q−r)+yq(r−p)+zr(p−q)=[a+(p−1)d2](q−r)+[a+(q−1)d2](r−p)+[a+(r−1)d2](p−q)\dfrac{x}{p}(q-r)+\dfrac{y}{q}(r-p)+\dfrac{z}{r}(p-q)=\left[ a+(p-1)\dfrac{d}{2} \right](q-r)+\left[ a+(q-1)\dfrac{d}{2} \right](r-p)+\left[ a+(r-1)\dfrac{d}{2} \right](p-q)
=a(q−r)+(p−1)d2(q−r)+a(r−p)+(q−1)d2(r−p)+a(p−q)+(r−1)d2(p−q)=a(q-r)+(p-1)\dfrac{d}{2}(q-r)+a(r-p)+(q-1)\dfrac{d}{2}(r-p)+a(p-q)+(r-1)\dfrac{d}{2}(p-q)
Combining like terms together we get,
=[a(q−r)+a(r−p)+a(p−q)]+[(p−1)d2(q−r)+(q−1)d2(r−p)+(r−1)d2(p−q)]=\left[ a(q-r)+a(r-p)+a(p-q) \right]+\left[ (p-1)\dfrac{d}{2}(q-r)+(q-1)\dfrac{d}{2}(r-p)+(r-1)\dfrac{d}{2}(p-q) \right]
Taking aa common from first bracket and d2\dfrac{d}{2} common from second bracket we get,
=a[q−r+r−p+p−q]+d2[(p−1)(q−r)+(q−1)(r−p)+(r−1)(p−q)]=a\left[ q-r+r-p+p-q \right]+\dfrac{d}{2}\left[ (p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q) \right]
=a[q−r+r−p+p−q]+d2[pq−pr−q+r+qr−qp−r+p+rp−rq−p+q]=a\left[ q-r+r-p+p-q \right]+\dfrac{d}{2}\left[ pq-pr-q+r+qr-qp-r+p+rp-rq-p+q \right]
The first bracket becomes zero and the second bracket also becomes zero, so
=a[0]+d2[0] =0 \begin{aligned} & =a\left[ 0 \right]+\dfrac{d}{2}\left[ 0 \right] \\\ & =0 \\\ \end{aligned}

Thus the value of xp(q−r)+yq(r−p)+zr(p−q)\dfrac{x}{p}(q-r)+\dfrac{y}{q}(r-p)+\dfrac{z}{r}(p-q) is zero.

Note: Sum of n terms in A.P is given by Sn=n2[2a+(n−1)d]{{S}_{n}}=\dfrac{n}{2}[2a+(n-1)d] ; where aa is the first term of A.P and d is the common difference. Care should be taken when we substitute the respective values of xp,yq,zr\dfrac{x}{p},\dfrac{y}{q},\dfrac{z}{r} inxp(q−r)+yq(r−p)+zr(p−q)\dfrac{x}{p}(q-r)+\dfrac{y}{q}(r-p)+\dfrac{z}{r}(p-q) plus and minus sign should be written carefully.