Question

Please explain step by step and answer

Answer 1

Answer:

The expression $a(q-r)+b(r-p)+c(p-q)$ equals to 0.

Step-by-step explanation:

The nth term of an AP is: $T_{n}=x+(n-1)d$

Here, x = first term and d = common difference.

Then the pth, qth and rth term are:

$T_{p} = x + (p - 1)d=a\\T_{q} = x + (q- 1)d=b\\T_{r} = x + (r - 1)d=c$

Substitute the value of a, b, and c in the provided expression and solve.

$a(q-r)+b(r-p)+c(p-q)\\=(x+(p-1)d)(q-r)+(x+(q-1)d)(r-p)+(x+(r-1)d)(p-q)\\=qx-rx+d(p-1)(q-r)+rx-px+d(q-1)(r-p)+px-qx+d(r-1)(p-q)\\=d(p-1)(q-r)+d(q-1)(r-p)+d(r-1)(p-q)\\=pq-pr-q+r+qr-qp-r+p+pr-qr-p+q\\=0$

Hence proved.

Answer 2

Answer:

ap=a

aq=b

ar=c(given)

Step-by-step explanation:

a+(p-1)d=a..................(1)

a+(q-1)d=b...................(2)

a+(r-1)d=c................(3)

by taking eq.1and 2

a+(p-1)d=a

a+(q-1)d=b

-. -. -

___________

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

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

(p-q)d=a-b

d=a-b÷p-q........(4)

by taking eq 2and 3

a+(q-1)d=b

a+(r-1)d=c

-. -. -

___________

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

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

(q-r)d=b-c

d=b-c÷q-r........(5)

from eq 4 and 5

a-b÷p-q=b-c÷q-r

by cross multiple

a-b(q-r)=b-c(p-q)

a(q-r)-b(q-r)=b(p-q)-c(p-q)

a(q-r)-b(q-r)-b(p-q)+c(p-q)

a(q-r)-bq+br-bp+bq+c(p-q)

a(q-r)+br-bp+c(p-q)

a(q-r)+b(r-p)+c(p-q)

hence proved

so simple