IF-THE-SUM-OF-p-TERMS-OF-AN-A-P-IS-EQUAL-TO-SUM-OF-ITS-q-TERMS-PROVE-THAT-THE-SUM-OF-p-q-TERMS-OF-IT-IS-EQUAL-TO-0-ZERO-

Question Number 80159 by Mr. AR last updated on 31/Jan/20

$$\boldsymbol{{IF}}\:\:\boldsymbol{{THE}}\:\:\:\boldsymbol{{SUM}}\:\:\:\boldsymbol{{OF}}\:\:\:\boldsymbol{{p}}\:\:\boldsymbol{{TERMS}} \\ $$$$\boldsymbol{{OF}}\:\:\boldsymbol{{AN}}\:\:\:\:\boldsymbol{{A}}.\boldsymbol{{P}}.\:\:\:\boldsymbol{{IS}}\:\:\:\boldsymbol{{EQUAL}}\:\:\boldsymbol{{TO}} \\ $$$$\boldsymbol{{SUM}}\:\:\boldsymbol{{OF}}\:\:\:\boldsymbol{{ITS}}\:\:\:\boldsymbol{{q}}\:\:\:\boldsymbol{{TERMS}}.\:\: \\ $$$$\boldsymbol{{PROVE}}\:\:\boldsymbol{{THAT}}\:\:\boldsymbol{{THE}}\:\:\boldsymbol{{SUM}}\:\:\boldsymbol{{OF}} \\ $$$$\left(\boldsymbol{{p}}+\boldsymbol{{q}}\right)\:\:\boldsymbol{{TERMS}}\:\:\boldsymbol{{OF}}\:\:\:\boldsymbol{{IT}}\:\:\:\boldsymbol{{IS}}\:\:\: \\ $$$$\boldsymbol{{EQUAL}}\:\:\boldsymbol{{TO}}\:\:\mathrm{0}\left(\boldsymbol{{ZERO}}\right). \\ $$

Commented by mr W last updated on 31/Jan/20

$${seems}\:{illogical} \\ $$

Commented by Rio Michael last updated on 31/Jan/20

sir sum of p terms of an AP and sum of q terms don′t actually make sense esspecially when the sum of (p + q) terms are involved.

$$\mathrm{sir}\:\mathrm{sum}\:\mathrm{of}\:{p}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{an}\:\mathrm{AP}\: \\ $$$$\mathrm{and}\:\mathrm{sum}\:\mathrm{of}\:{q}\:\mathrm{terms}\:\mathrm{don}'\mathrm{t}\:\mathrm{actually}\:\mathrm{make} \\ $$$$\mathrm{sense}\:\mathrm{esspecially}\:\mathrm{when}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\: \\ $$$$\left({p}\:+\:{q}\right)\:\mathrm{terms}\:\mathrm{are}\:\mathrm{involved}. \\ $$

Commented by $@ty@m123 last updated on 31/Jan/20

Example: Consider the following A.P. 1,(1/2),0,((−1)/2),−1 Here, S_2 =S_3 & S_5 =0

$${Example}: \\ $$$${Consider}\:{the}\:{following}\:{A}.{P}.\: \\ $$$$\mathrm{1},\frac{\mathrm{1}}{\mathrm{2}},\mathrm{0},\frac{−\mathrm{1}}{\mathrm{2}},−\mathrm{1} \\ $$$${Here}, \\ $$$${S}_{\mathrm{2}} ={S}_{\mathrm{3}} \\ $$$$\&\:{S}_{\mathrm{5}} =\mathrm{0} \\ $$

Commented by mr W last updated on 01/Feb/20

$${the}\:{question}\:{didn}'{t}\:{say}\:{the}\:{first}\:{p}\:{terms}, \\ $$$${the}\:{first}\:{q}\:{terms}\:{and}\:{the}\:{first}\:{p}+{q} \\ $$$${terms}. \\ $$

Answered by $@ty@m123 last updated on 31/Jan/20

S_p =(p/2){2a+(p−1)d} S_q =(q/2){2a+(q−1)d} Given, S_p =S_q (p/2){2a+(p−1)d}=(q/2){2a+(q−1)d} p{2a+(p−1)d}=q{2a+(q−1)d} p{2a+(p−1)d}−q{2a+(q−1)d}=0 2a(p−q)+d{p(p−1)−q(q−1)}=0 2a(p−q)+d(p^2 −p−q^2 +q)=0 2a(p−q)+d{(p^2 −q^2 )−(p−q)}=0 2a(p−q)+d(p−q)(p+q−1)=0 (p−q){2a+(p+q−1)d}=0 {2a+(p+q−1)d}=0∨p=q {2a+(p+q−1)d}=0 ⇒ ((p+q)/2){2a+(p+q−1)d}=0 S_(p+q) =0

$${S}_{{p}} =\frac{{p}}{\mathrm{2}}\left\{\mathrm{2}{a}+\left({p}−\mathrm{1}\right){d}\right\} \\ $$$${S}_{{q}} =\frac{{q}}{\mathrm{2}}\left\{\mathrm{2}{a}+\left({q}−\mathrm{1}\right){d}\right\} \\ $$$${Given}, \\ $$$${S}_{{p}} ={S}_{{q}} \\ $$$$\frac{{p}}{\mathrm{2}}\left\{\mathrm{2}{a}+\left({p}−\mathrm{1}\right){d}\right\}=\frac{{q}}{\mathrm{2}}\left\{\mathrm{2}{a}+\left({q}−\mathrm{1}\right){d}\right\} \\ $$$${p}\left\{\mathrm{2}{a}+\left({p}−\mathrm{1}\right){d}\right\}={q}\left\{\mathrm{2}{a}+\left({q}−\mathrm{1}\right){d}\right\} \\ $$$${p}\left\{\mathrm{2}{a}+\left({p}−\mathrm{1}\right){d}\right\}−{q}\left\{\mathrm{2}{a}+\left({q}−\mathrm{1}\right){d}\right\}=\mathrm{0} \\ $$$$\mathrm{2}{a}\left({p}−{q}\right)+{d}\left\{{p}\left({p}−\mathrm{1}\right)−{q}\left({q}−\mathrm{1}\right)\right\}=\mathrm{0} \\ $$$$\mathrm{2}{a}\left({p}−{q}\right)+{d}\left({p}^{\mathrm{2}} −{p}−{q}^{\mathrm{2}} +{q}\right)=\mathrm{0} \\ $$$$\mathrm{2}{a}\left({p}−{q}\right)+{d}\left\{\left({p}^{\mathrm{2}} −{q}^{\mathrm{2}} \right)−\left({p}−{q}\right)\right\}=\mathrm{0} \\ $$$$\mathrm{2}{a}\left({p}−{q}\right)+{d}\left({p}−{q}\right)\left({p}+{q}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({p}−{q}\right)\left\{\mathrm{2}{a}+\left({p}+{q}−\mathrm{1}\right){d}\right\}=\mathrm{0} \\ $$$$\left\{\mathrm{2}{a}+\left({p}+{q}−\mathrm{1}\right){d}\right\}=\mathrm{0}\vee{p}={q} \\ $$$$\left\{\mathrm{2}{a}+\left({p}+{q}−\mathrm{1}\right){d}\right\}=\mathrm{0}\:\Rightarrow \\ $$$$\frac{{p}+{q}}{\mathrm{2}}\left\{\mathrm{2}{a}+\left({p}+{q}−\mathrm{1}\right){d}\right\}=\mathrm{0} \\ $$$${S}_{{p}+{q}} =\mathrm{0} \\ $$

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IF-THE-SUM-OF-p-TERMS-OF-AN-A-P-IS-EQUAL-TO-SUM-OF-ITS-q-TERMS-PROVE-THAT-THE-SUM-OF-p-q-TERMS-OF-IT-IS-EQUAL-TO-0-ZERO-

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