Question

if p,q and r term of AP be a,b,c respectively than show that
a(q–r) + b(r–p) + c(p–q)=0​

Answer 1

Answer:

Step-by-step explanation:

Let A be the first term and d be the common difference of the AP

Then,

pth term=a=A+(p-1)d

qth term=b=A+(q-1)d

rth term =c=A+(r-q)d

We have,

a(q–r) + b(r–p) + c(p–q) =>

{A+(p-1)d}(q–r) + {A+(q-1)d}(r-p) + {A+(r-q)d}(p-q)=

Opening up all the paranthesis and simplyfying we get that the expression is equal to 0 (Try this yourself)

∴ a(q–r) + b(r–p) + c(p–q)=0​

Answer 2

Answer:

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