Question

if p,q,r,s are in GP then show that (q-r)^2+ ( r-p)^2 + (s-q)^2 = (p-s)^2​

Answer 1

Answer:

Let common ratio = k

so q = pk, r = pk², s = pk³

LHS

${(q - r)}^{2} + {(r - p)}^{2} + {(s - q)}^{2} =$

${(pk - p {k}^{2} )}^{2} + {(p {k}^{2} - p) }^{2} + {(pk {}^{3} - pk)}^{2} =$

${p}^{2} {k}^{2} {(1 - k)}^{2} + {p}^{2} (k {}^{2} - 1) {}^{2} + {p}^{2} {k}^{2} ( {k}^{2} - 1) {}^{2} = (k - 1) {}^{2} ( {p}^{2} {k}^{2} + {p}^{2} {(k + 1)}^{2} + {p}^{2} {k}^{2} {(k + 1)}^{2} )$