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In-an-AP-if-the-m-th-term-is-n-and-the-n-th-term-is-m-where-m-is-not-equal-to-n-find-1-The-value-of-m-n-th-term-2-The-p-th-term-

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Question Number 7399 by Tawakalitu. last updated on 27/Aug/16
In an AP , if the m^(th) term is n and the n^(th) term is m. where m is not equal to n. find (1) The value of (m + n)^(th) term (2) The p^(th) term.
$${In}\:{an}\:{AP}\:,\:{if}\:{the}\:{m}^{{th}} \:{term}\:{is}\:{n}\:{and}\:{the}\:{n}^{{th}} \:{term} \\ $$$${is}\:{m}.\:{where}\:{m}\:{is}\:{not}\:{equal}\:{to}\:{n}.\:{find} \\ $$$$\left(\mathrm{1}\right)\:{The}\:{value}\:{of}\:\left({m}\:+\:{n}\right)^{{th}} \:{term} \\ $$$$\left(\mathrm{2}\right)\:{The}\:{p}^{{th}} \:{term}. \\ $$
Answered by Yozzia last updated on 27/Aug/16
For an AP, the k^(th) term u(k) is given by u(k)=u(1)+(k−1)d where d is a constant. ∴ k=m⇒u(m)=u(1)+(m−1)d=n and k=n⇒u(n)=u(1)+(n−1)d=m (1) k=m+n⇒u(m+n)=u(1)+(m+n−1)d But, n−m=(m−n)d and m≠n⇒d=−1 ∴u(m+n)=u(1)+(m+n−1)(−1) but, n+m=2u(1)+(−1)(m+n−2) 2(m+n)=2u(1)+2 u(1)=m+n−1 ∴ u(m+n)=m+n−1+(−1)(m+n−1)=0 (2) k=p⇒ u(p)=m+n−1+(p−1)(−1)=m+n−p
$${For}\:{an}\:{AP},\:{the}\:{k}^{{th}} \:{term}\:{u}\left({k}\right)\:{is}\:{given}\:{by} \\ $$$${u}\left({k}\right)={u}\left(\mathrm{1}\right)+\left({k}−\mathrm{1}\right){d}\:{where}\:{d}\:{is}\:{a}\:{constant}. \\ $$$$\therefore\:{k}={m}\Rightarrow{u}\left({m}\right)={u}\left(\mathrm{1}\right)+\left({m}−\mathrm{1}\right){d}={n} \\ $$$${and}\:{k}={n}\Rightarrow{u}\left({n}\right)={u}\left(\mathrm{1}\right)+\left({n}−\mathrm{1}\right){d}={m} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:{k}={m}+{n}\Rightarrow{u}\left({m}+{n}\right)={u}\left(\mathrm{1}\right)+\left({m}+{n}−\mathrm{1}\right){d} \\ $$$${But},\:{n}−{m}=\left({m}−{n}\right){d}\:\:{and}\:{m}\neq{n}\Rightarrow{d}=−\mathrm{1} \\ $$$$\therefore{u}\left({m}+{n}\right)={u}\left(\mathrm{1}\right)+\left({m}+{n}−\mathrm{1}\right)\left(−\mathrm{1}\right) \\ $$$${but},\:{n}+{m}=\mathrm{2}{u}\left(\mathrm{1}\right)+\left(−\mathrm{1}\right)\left({m}+{n}−\mathrm{2}\right) \\ $$$$\mathrm{2}\left({m}+{n}\right)=\mathrm{2}{u}\left(\mathrm{1}\right)+\mathrm{2} \\ $$$${u}\left(\mathrm{1}\right)={m}+{n}−\mathrm{1} \\ $$$$\therefore\:{u}\left({m}+{n}\right)={m}+{n}−\mathrm{1}+\left(−\mathrm{1}\right)\left({m}+{n}−\mathrm{1}\right)=\mathrm{0} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:{k}={p}\Rightarrow\:{u}\left({p}\right)={m}+{n}−\mathrm{1}+\left({p}−\mathrm{1}\right)\left(−\mathrm{1}\right)={m}+{n}−{p} \\ $$
Commented by Rasheed Soomro last updated on 27/Aug/16
Another way From the above d=−1 u(1)+(m−1)d=n adding nd to both sides u(1)+(m−1)d+nd=n+nd u(1)+(m+n−1)d=n+n(−1) u(1)+(m+n−1)d=0 u(m+n)=0 u(1)+(m−1)d=n Subtracting (m−p)d to both sides u(1)+(m−1)d−(m−p)d=n−(m−p)d u(1)+(m−1−m+p)d=n−(m−p)(−1) u(1)+(p−1)d=n+m−p u(p)=m+n−p
$${Another}\:{way} \\ $$$${From}\:{the}\:{above}\:\:{d}=−\mathrm{1} \\ $$$${u}\left(\mathrm{1}\right)+\left({m}−\mathrm{1}\right){d}={n} \\ $$$${adding}\:{nd}\:{to}\:{both}\:{sides} \\ $$$${u}\left(\mathrm{1}\right)+\left({m}−\mathrm{1}\right){d}+{nd}={n}+{nd} \\ $$$${u}\left(\mathrm{1}\right)+\left({m}+{n}−\mathrm{1}\right){d}={n}+{n}\left(−\mathrm{1}\right) \\ $$$${u}\left(\mathrm{1}\right)+\left({m}+{n}−\mathrm{1}\right){d}=\mathrm{0} \\ $$$${u}\left({m}+{n}\right)=\mathrm{0} \\ $$$$ \\ $$$${u}\left(\mathrm{1}\right)+\left({m}−\mathrm{1}\right){d}={n} \\ $$$${Subtracting}\:\left({m}−{p}\right){d}\:{to}\:{both}\:{sides} \\ $$$${u}\left(\mathrm{1}\right)+\left({m}−\mathrm{1}\right){d}−\left({m}−{p}\right){d}={n}−\left({m}−{p}\right){d} \\ $$$${u}\left(\mathrm{1}\right)+\left({m}−\mathrm{1}−{m}+{p}\right){d}={n}−\left({m}−{p}\right)\left(−\mathrm{1}\right) \\ $$$${u}\left(\mathrm{1}\right)+\left({p}−\mathrm{1}\right){d}={n}+{m}−{p} \\ $$$${u}\left({p}\right)={m}+{n}−{p} \\ $$
Commented by Tawakalitu. last updated on 27/Aug/16
Thank you sir. i really appreciate
$${Thank}\:{you}\:{sir}.\:{i}\:{really}\:{appreciate} \\ $$

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