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15. The pth, qth and rth terms of an A.P. are a, b, c, respectively. Show that

$(q - r )a + (r - p )b + (p - q )c = 0$

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Given: The pth, qth and rth terms of an A.P. are a, b, c, respectively.

To prove : $(q - r )a + (r - p )b + (p - q )c = 0$

Let the first term of AP be 't' and common difference be d

$a_p=t+(p-1)d=a...............................1$

$a_q=t+(q-1)d=b...............................2$

$a_r=t+(r-1)d=c...............................3$

Subtracting equation 2 from 1, we get

$(p-1-q+1)d=a-b$

$\Rightarrow (p-q)d=a-b$

$\Rightarrow d=\frac{a-b}{p-q}....................................4$

Subtracting equation 3 from 2, we get

$(q-1-r+1)d=b-c$

$\Rightarrow (q-r)d=b-c$

$\Rightarrow d=\frac{b-c}{q-r}....................................5$

Equating values of d, from equation 4 and 5, we have

$d=\frac{a-b}{p-q}=\frac{b-c}{q-r}.$

$\Rightarrow \frac{a-b}{p-q}=\frac{b-c}{q-r}.$

$\Rightarrow (a-b)(q-r)=(b-c)(p-q)$

$\Rightarrow aq-ar-bq+br=bp-bq-cp+cq$

$\Rightarrow aq-ar+br=bp-cp+cq$

$\Rightarrow aq-ar+br-bp+cp-cq=0$

$\Rightarrow a(q-r)+b(r-p)+c(p-q)=0$

Hence proved.

Posted by

seema garhwal

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