Question

ii) If pth, qth and th terms of an A.P are x, y, z respectively, show that
(9-r) x + (r-p)y+ (p-q)2 = 0​

Answer 1

Correct Question:

If the pth, qth and rth terms of an AP be a, b, c respectively, then show that  a(q - r) + b(r - p) + c(p - q) = 0.

Answer:

Hence Proved!!

Step-by-step explanation:

Given:

• If the pth, qth and rth terms of an AP be x, y and z.

To Show:

• a(q - r) + b(r - p)+ c(p - q) = 0

Let the first term of an AP is 'a' and common difference is 'd'.

The pth, qth and rth term of an A.P. are x, y and z respectively. So,

$\longrightarrow \sf x=a+(p-1)d\;\;\;\;\;\;.........(1)\\ \\ \\ \longrightarrow \sf y=a+(q-1)d\;\;\;\;\;\;.........(2)\\ \\ \\ \longrightarrow \sf z=a+(r-1)d\;\;\;\;\;\;.........(3)$

Now, subtracting Equation (1) from Equation (2), we get,

$\longrightarrow \sf (x-y)=[(p-1)-(q-1)]d\\ \\ \\ \longrightarrow \sf (x-y)=(p-q)d\\ \\ \\ \longrightarrow d=\dfrac{x-y}{p-q}$

Now, Substitute value of d in equation (1), we get,

$\longrightarrow \sf x=a+(p-1)\Bigg(\dfrac{x-y}{p-q}\Bigg)\\ \\ \\ \longrightarrow \sf a=x-(p-1)\Bigg(\dfrac{x-y}{p-q}\Bigg)$

Now, Substitute value of m and d in equation (3), we get,

$\longrightarrow \sf z=\Bigg[x-(p-1)\bigg(\dfrac{x-y}{p-q}\Bigg)\Bigg]+(r-1)\Bigg(\dfrac{x-y}{p-q}\Bigg)\\ \\ \\ \longrightarrow \sf z=\dfrac{x(p-q)-(p-1)(x-y)+(r-1)(x-y)}{(p-q)}\\ \\ \\ \longrightarrow \sf z=\dfrac{xp-xp-(xp-py-x+y)+xr-ry-x+y}{(p-q)}\\ \\ \\ \longrightarrow \sf z=\dfrac{-xq+py+xr-ry}{(p-q)}\\ \\ \\ \longrightarrow \sf z = \dfrac{x(r-q)+y(p-r)}{(p-q)}\\ \\ \\ \longrightarrow \sf z(p-q)=x(r-q) +y(p-r)\\ \\ \\ \longrightarrow \sf x(q-r)+y(r-p)+z(p-q)=0$

Hence Proved!!

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#BAL