If p times the pth term of an A.P is equal to q times the qth term , find (p+q)th term of A.P
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Given:
p*Tp = q*Tq
Using Tn formula,
Tp = a + (p-1)d
Tq = a+(q-1)d
p*Tp = p[a+(p-1)d] = ap +pd(p-1) ----------- (1)
q*Tq = q[a+(q-1)d] = aq + qd(q-1) ------------(2)
equating (1) and (2)
ap + pd(p-1) = aq + qd(q-1)
ap -aq = qd(q-1) - pd(p-1)
a(p-q) = d[(q^2-q-p^2+p]
a(p-q) = d[q^2-p^2 - (q-p)]
a(p-q) = d[(q-p)(q+p)-(q-p)]
a(p-q) = d(q-p)[q+p-1]
a=d(q-p)[q+p-1]/(p-q)
a=-d(q+p-1)
Tp+q = a +(p+q-1)d
Substituting a in the above formula, Tp+q =- d(q+p-1) +(p+q-1)d =0
Using Tn formula,
Tp = a + (p-1)d
Tq = a+(q-1)d
p*Tp = p[a+(p-1)d] = ap +pd(p-1) ----------- (1)
q*Tq = q[a+(q-1)d] = aq + qd(q-1) ------------(2)
equating (1) and (2)
ap + pd(p-1) = aq + qd(q-1)
ap -aq = qd(q-1) - pd(p-1)
a(p-q) = d[(q^2-q-p^2+p]
a(p-q) = d[q^2-p^2 - (q-p)]
a(p-q) = d[(q-p)(q+p)-(q-p)]
a(p-q) = d(q-p)[q+p-1]
a=d(q-p)[q+p-1]/(p-q)
a=-d(q+p-1)
Tp+q = a +(p+q-1)d
Substituting a in the above formula, Tp+q =- d(q+p-1) +(p+q-1)d =0
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