Question

The sum of the first p, q, r, terms of an AP are a, b, c, respectively. Show that (q-r)+(r-p)+(p-q) =0

Answer 1

ANWER.


✷GIVEN

pth, qth , rth terms of an AP Are a, b,c respectively.


✷PROVE

Show that (q-r)+(r-p)+(p-q) =0



✸SOLUTION.


☞FORMULA USED

nth term formula


a(n) = a+(n+1)d


so.

✍STEP. 1

For. pth term the eq will be


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



For qth term the eq will be



a(q) = a+(q+1)d. ........................eq(2)




For rth term the eq will be




a(r)= a+(r+1)d. .............................eq(3)

STEP2


✍NOW USING GIVEN INFORMATION.



The all eq are equal to


eq(1) is equal to a. but I use A instead of a
because it comes twice in question in formula and given information also


so let's start



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



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



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


☞SOLUTION OF THESE EQ.

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


=>a+(p+1)d. =A. ...........eq(4)


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

=>a+(q+1)d= b. ..........eq (5)


now solve both substracted both eq

we,get. ( p+1)d -(q+1)d=A - b

(pd+d -qd - d)=A - b

d(p - q)=A - b. .............eq(6)

now put a inplace of A
$(p - q) = \frac{a - b}{d}$


similarly


$(r - p) = \frac{c - a}{d}$


$(q - r) = \frac{b - c}{d}$


now add we get


$(p - q) + (r - p) + (q - r) =$


$= \frac{a - b}{d} + \frac{c - a}{d} + \frac{b - c}{d}$


$= \frac{( a - b) + (c - a) + (b- c)}{d}$


$\frac{0}{d}$


0



HENCE PROVED