Question

Solve this sum.........if you can..

Answer 1

hope it helps..... :)

Answer 2

Hey there,

Answer:

(pr)+(rp)=(ac)+(ca)

See proof below.

Explanation:

If a, b, c are in A.P; we have 2b=a+c -------- (A)

If ap, bq, cr are in G.P; then (bq)2=ap×cr or rpq2=b2ac ------- (B)

and as p, q, r are in H.P, 1p, 1q, 1r are in A.P. and 
2q=1p+1r=p+rrp or (p+r)=2rpq ---------- (C)

Now (pr)+(rp)=p2+r2rp=(p+r)2−2prrp=(p+r)2rp−2

Using relation (C) for (p+r), we get

(pr)+(rp)=(2rpq)2×1rp−2 or

(pr)+(rp)=(4rpq2)−2 and using (B) this becomes

(pr)+(rp)=(4b2ac)−2=(2b)2ac−2=(a+c)2ac−2 i.e.

(pr)+(rp)=(a+c)2−2acac=a2+c2ac or

(pr)+(rp)=(ac)+(ca)

Hope this helps!