Question Number 46922 by 786786AM last updated on 02/Nov/18
$$\mathrm{Thr}\:\mathrm{6th}\:\mathrm{term}\:\mathrm{of}\:\mathrm{an}\:\mathrm{AP}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{2},\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{common}\:\mathrm{difference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{AP}. \\ $$$$\mathrm{Which}\:\mathrm{makes}\:\mathrm{the}\:\mathrm{product}\:{a}_{\mathrm{1}} \:{a}_{\mathrm{4}} \:{a}_{\mathrm{5}} \:\mathrm{least} \\ $$$$\mathrm{is}\:\mathrm{given}\:\:\mathrm{by} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 03/Nov/18
![a+5d=2 p=a(a+3d)(a+4d) p=(2−5d)(2−5d+3d)(2−5d+4d) p=(2−5d)(2−2d)(2−d) lnp=ln(2−5d)+ln(2−2d)+ln(2−d) (1/p)×(dp/(d(d)))=(1/(2−5d))×(−5)+(1/(2−2d))×(−2)+(1/(2−d))×(−1) Dp=p×[((−5)/(2−5d))+((−2)/(2−2d))+((−1)/(2−d))] =−5(2−2d)(2−d)−2(2−5d)(2−d)−(2−5d)(2−2d) =−5(4−6d+2d^2 )−2(4−12d+5d^2 )−1(4−14d+10d^2 ) for max/min Dp=0 20−30d+10d^2 +8−24d+10d^2 +4−14d+10d^2 =30d^2 −68d+32 15d^2 −34d+16=0 d=((34±(√(34^2 −4×15×16)))/(2×15)) d=((34±(√(1156−960)))/(30)) d=((34±14)/(30)) (((48)/(30)),((20)/(30)))→((8/5),(2/3)) Dp=30d^2 −68d+16 D^2 p=60d−68 value of D^2 p at d=(8/5) is 60×(8/5)−68=96−68=28 +ve value at (2/3)is 60×(2/3)−68=40−68=−28 −ve so at d=(2/3) the product is maximum and d=(8/5) it is minimum... now a+5d=2 a+5×(8/5)=2 a=−6 a_1 ×a_4 ×a_5 (−6)×(−6+3×(8/5))(−6+4×(8/5)) =(−6)×(((−30+24)/5))(((−30+32)/5)) =(−6)×(((−6)/5))((2/5))=((72)/(25)) pls check....](https://www.tinkutara.com/question/Q46968.png)
$${a}+\mathrm{5}{d}=\mathrm{2} \\ $$$${p}={a}\left({a}+\mathrm{3}{d}\right)\left({a}+\mathrm{4}{d}\right) \\ $$$${p}=\left(\mathrm{2}−\mathrm{5}{d}\right)\left(\mathrm{2}−\mathrm{5}{d}+\mathrm{3}{d}\right)\left(\mathrm{2}−\mathrm{5}{d}+\mathrm{4}{d}\right) \\ $$$${p}=\left(\mathrm{2}−\mathrm{5}{d}\right)\left(\mathrm{2}−\mathrm{2}{d}\right)\left(\mathrm{2}−{d}\right) \\ $$$${lnp}={ln}\left(\mathrm{2}−\mathrm{5}{d}\right)+{ln}\left(\mathrm{2}−\mathrm{2}{d}\right)+{ln}\left(\mathrm{2}−{d}\right) \\ $$$$\frac{\mathrm{1}}{{p}}×\frac{{dp}}{{d}\left({d}\right)}=\frac{\mathrm{1}}{\mathrm{2}−\mathrm{5}{d}}×\left(−\mathrm{5}\right)+\frac{\mathrm{1}}{\mathrm{2}−\mathrm{2}{d}}×\left(−\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{2}−{d}}×\left(−\mathrm{1}\right) \\ $$$${Dp}={p}×\left[\frac{−\mathrm{5}}{\mathrm{2}−\mathrm{5}{d}}+\frac{−\mathrm{2}}{\mathrm{2}−\mathrm{2}{d}}+\frac{−\mathrm{1}}{\mathrm{2}−{d}}\right] \\ $$$$=−\mathrm{5}\left(\mathrm{2}−\mathrm{2}{d}\right)\left(\mathrm{2}−{d}\right)−\mathrm{2}\left(\mathrm{2}−\mathrm{5}{d}\right)\left(\mathrm{2}−{d}\right)−\left(\mathrm{2}−\mathrm{5}{d}\right)\left(\mathrm{2}−\mathrm{2}{d}\right) \\ $$$$=−\mathrm{5}\left(\mathrm{4}−\mathrm{6}{d}+\mathrm{2}{d}^{\mathrm{2}} \right)−\mathrm{2}\left(\mathrm{4}−\mathrm{12}{d}+\mathrm{5}{d}^{\mathrm{2}} \right)−\mathrm{1}\left(\mathrm{4}−\mathrm{14}{d}+\mathrm{10}{d}^{\mathrm{2}} \right) \\ $$$${for}\:{max}/{min}\:{Dp}=\mathrm{0} \\ $$$$\mathrm{20}−\mathrm{30}{d}+\mathrm{10}{d}^{\mathrm{2}} +\mathrm{8}−\mathrm{24}{d}+\mathrm{10}{d}^{\mathrm{2}} +\mathrm{4}−\mathrm{14}{d}+\mathrm{10}{d}^{\mathrm{2}} \\ $$$$=\mathrm{30}{d}^{\mathrm{2}} −\mathrm{68}{d}+\mathrm{32} \\ $$$$\mathrm{15}{d}^{\mathrm{2}} −\mathrm{34}{d}+\mathrm{16}=\mathrm{0} \\ $$$${d}=\frac{\mathrm{34}\pm\sqrt{\mathrm{34}^{\mathrm{2}} −\mathrm{4}×\mathrm{15}×\mathrm{16}}}{\mathrm{2}×\mathrm{15}} \\ $$$${d}=\frac{\mathrm{34}\pm\sqrt{\mathrm{1156}−\mathrm{960}}}{\mathrm{30}} \\ $$$${d}=\frac{\mathrm{34}\pm\mathrm{14}}{\mathrm{30}}\:\:\left(\frac{\mathrm{48}}{\mathrm{30}},\frac{\mathrm{20}}{\mathrm{30}}\right)\rightarrow\left(\frac{\mathrm{8}}{\mathrm{5}},\frac{\mathrm{2}}{\mathrm{3}}\right) \\ $$$${Dp}=\mathrm{30}{d}^{\mathrm{2}} −\mathrm{68}{d}+\mathrm{16} \\ $$$${D}^{\mathrm{2}} {p}=\mathrm{60}{d}−\mathrm{68} \\ $$$${value}\:{of}\:{D}^{\mathrm{2}} {p}\:{at}\:{d}=\frac{\mathrm{8}}{\mathrm{5}}\:{is} \\ $$$$\mathrm{60}×\frac{\mathrm{8}}{\mathrm{5}}−\mathrm{68}=\mathrm{96}−\mathrm{68}=\mathrm{28}\:\:\:\:+{ve} \\ $$$${value}\:{at}\:\frac{\mathrm{2}}{\mathrm{3}}{is}\: \\ $$$$\mathrm{60}×\frac{\mathrm{2}}{\mathrm{3}}−\mathrm{68}=\mathrm{40}−\mathrm{68}=−\mathrm{28}\:−{ve} \\ $$$${so}\:{at}\:{d}=\frac{\mathrm{2}}{\mathrm{3}}\:\:{the}\:{product}\:{is}\:{maximum}\:{and}\:{d}=\frac{\mathrm{8}}{\mathrm{5}} \\ $$$${it}\:{is}\:{minimum}… \\ $$$${now}\:{a}+\mathrm{5}{d}=\mathrm{2} \\ $$$${a}+\mathrm{5}×\frac{\mathrm{8}}{\mathrm{5}}=\mathrm{2}\:\:\:{a}=−\mathrm{6} \\ $$$${a}_{\mathrm{1}} ×{a}_{\mathrm{4}} ×{a}_{\mathrm{5}} \\ $$$$\left(−\mathrm{6}\right)×\left(−\mathrm{6}+\mathrm{3}×\frac{\mathrm{8}}{\mathrm{5}}\right)\left(−\mathrm{6}+\mathrm{4}×\frac{\mathrm{8}}{\mathrm{5}}\right) \\ $$$$=\left(−\mathrm{6}\right)×\left(\frac{−\mathrm{30}+\mathrm{24}}{\mathrm{5}}\right)\left(\frac{−\mathrm{30}+\mathrm{32}}{\mathrm{5}}\right) \\ $$$$=\left(−\mathrm{6}\right)×\left(\frac{−\mathrm{6}}{\mathrm{5}}\right)\left(\frac{\mathrm{2}}{\mathrm{5}}\right)=\frac{\mathrm{72}}{\mathrm{25}}\:\:{pls}\:{check}…. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$