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Worked Examples · Example 5

Q.If pp times the ppth term of an A.P. is equal to qq times the qqth term, then show that its (p+q)(p+q)th term is zero.

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Writing both terms with an=a+(n−1)da_n=a+(n-1)d and simplifying the given condition factors out (p−q)(p-q), forcing ap+q=0a_{p+q}=0.

For an A.P. with first term aa and common difference dd:

an=a+(n−1)da_n = a+(n-1)d

where ana_n is the nnth term. We are given p⋅ap=q⋅aqp\cdot a_p = q\cdot a_q and must show ap+q=0a_{p+q}=0.

  1. Write the ppth and qqth terms: ap=a+(p−1)da_p = a+(p-1)d and aq=a+(q−1)da_q = a+(q-1)d.
  2. Given condition: p ap=q aqp\,a_p = q\,a_q, i.e. p[a+(p−1)d]=q[a+(q−1)d]p[a+(p-1)d] = q[a+(q-1)d].
  3. Expand both sides: pa+p(p−1)d=qa+q(q−1)dpa+p(p-1)d = qa+q(q-1)d.
  4. Bring everything to one side: a(p−q)+d[p(p−1)−q(q−1)]=0a(p-q) + d\big[p(p-1)-q(q-1)\big] = 0.
  5. Simplify the bracket: p(p−1)−q(q−1)=(p2−q2)−(p−q)=(p−q)(p+q)−(p−q)=(p−q)(p+q−1)p(p-1)-q(q-1) = (p^2-q^2)-(p-q) = (p-q)(p+q)-(p-q) = (p-q)(p+q-1).
  6. Substitute back: a(p−q)+d(p−q)(p+q−1)=0  ⇒  (p−q)[a+d(p+q−1)]=0a(p-q) + d(p-q)(p+q-1) = 0 \;\Rightarrow\; (p-q)\big[a+d(p+q-1)\big] = 0. …

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