Question-188360

Question Number 188360 by 073 last updated on 28/Feb/23

Answered by Rasheed.Sindhi last updated on 28/Feb/23

S_N =(N/2)(a+l) A=((n−2)/2)(1+n−2)=(((n−2)(n−1))/2) B=((n−14)/2)(15+n)=(((n−14)(n+15))/2) A−B=(((n−2)(n−1))/2)−(((n−14)(n+15))/2)=42 =(((n−2)(n−1)−(n−14)(n+15))/2)=42 =(((n^2 −3n+2)−(n^2 +n−210))/2)=42 =((−4n+212)/2)=106−2n=42 n=32

$$\:\:\mathrm{S}_{\mathrm{N}} =\frac{\mathrm{N}}{\mathrm{2}}\left({a}+{l}\right) \\ $$$$\:\:\:\mathrm{A}=\frac{{n}−\mathrm{2}}{\mathrm{2}}\left(\mathrm{1}+{n}−\mathrm{2}\right)=\frac{\left({n}−\mathrm{2}\right)\left({n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\:\:\:\mathrm{B}=\frac{{n}−\mathrm{14}}{\mathrm{2}}\left(\mathrm{15}+{n}\right)=\frac{\left({n}−\mathrm{14}\right)\left({n}+\mathrm{15}\right)}{\mathrm{2}} \\ $$$$\mathrm{A}−\mathrm{B}=\frac{\left({n}−\mathrm{2}\right)\left({n}−\mathrm{1}\right)}{\mathrm{2}}−\frac{\left({n}−\mathrm{14}\right)\left({n}+\mathrm{15}\right)}{\mathrm{2}}=\mathrm{42} \\ $$$$\:\:=\frac{\left({n}−\mathrm{2}\right)\left({n}−\mathrm{1}\right)−\left({n}−\mathrm{14}\right)\left({n}+\mathrm{15}\right)}{\mathrm{2}}=\mathrm{42} \\ $$$$\:\:=\frac{\left({n}^{\mathrm{2}} −\mathrm{3}{n}+\mathrm{2}\right)−\left({n}^{\mathrm{2}} +{n}−\mathrm{210}\right)}{\mathrm{2}}=\mathrm{42} \\ $$$$\:\:=\frac{−\mathrm{4}{n}+\mathrm{212}}{\mathrm{2}}=\mathrm{106}−\mathrm{2}{n}=\mathrm{42} \\ $$$$\:\:\:\:{n}=\mathrm{32} \\ $$

Commented by 073 last updated on 01/Mar/23

$$\mathrm{thanks}\:\mathrm{sir} \\ $$

Answered by manxsol last updated on 28/Feb/23

A−B=Σ_1 ^(14) n−(n−1)−n 42=((14×15)/2)−2n+1 42=105−2n+1 2n=64 n=32

$${A}−{B}=\sum_{\mathrm{1}} ^{\mathrm{14}} {n}−\left({n}−\mathrm{1}\right)−{n} \\ $$$$\mathrm{42}=\frac{\mathrm{14}×\mathrm{15}}{\mathrm{2}}−\mathrm{2}{n}+\mathrm{1} \\ $$$$\mathrm{42}=\mathrm{105}−\mathrm{2}{n}+\mathrm{1} \\ $$$$\mathrm{2}{n}=\mathrm{64} \\ $$$${n}=\mathrm{32} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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