Question

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Class - 10th
Chapter - Arithmetic progression.

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Answer 1

Answer:

See the attachment which is given above for solution

Step-by-step explanation:

Used concept:-

• If a series of numbers are in the AP then the common difference must be same throughout the entire series.
• If a is the first term and d is the common difference then the nth term of the AP is denoted by an and it is defined by

an=a+(n-1) d.

Answer 2

Let a = first term of the AP.

and 

Let d = common difference of the AP

Now

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

b = A+(q-1).d.......(ii)

c = A+(r-1).d........(iii)

Subtracting 2nd from 1st , 3rd from 2nd and 1st from 3rd we get

a-b = (p-q).d......(iv)

b-c = (q-r).d........(v)

c-a = (r-p).d.......(vi)

multiply iv,v,vi by c,a,b respectively we have

c.(a-b) = c.(p-q).d......(vii)

a.(b-c) = a.(q-r).d........(viii)

b.(c-a) = b.(r-p).d.......(ix)

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

Now since d is common difference it should be non zero

Hence

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