Question Number 174871 by Ml last updated on 13/Aug/22
Answered by Rasheed.Sindhi last updated on 13/Aug/22
![a_(15) =32, S_9 +S_(29) =100, a_(23) =? Assuming a_n an AP a_(15) =a+(15−1)d=32 a_(15) =a+14d=32..................(i) S_9 +S_(29) =(9/2)(2a+8d)+((29)/2)(2a+28d)=100 9(a+4d)+29(a+14d)=100 9(a+4d)+29(32)=100 a+4d=((100−29×32)/9)=−92.....(ii) (i)−(ii): 10d=32+92=124 d=((124)/(10))=((62)/5) a+4(((62)/5))=−92⇒a=−92−((248)/5)=−((708)/5) a_(23) =a+22d=−((708)/5)+22(((62)/5))=((656)/5) ⟨•⟩FORMULAS USED⟨•⟩ determinant (((a_n =a+(n−1)d)),((S_n =(n/2)[ 2a+(n−1)d ])))](Q174881.png)
$${a}_{\mathrm{15}} =\mathrm{32},\:\:\:{S}_{\mathrm{9}} +{S}_{\mathrm{29}} =\mathrm{100},\:\:{a}_{\mathrm{23}} =? \\ $$$$\underline{{Assuming}\:{a}_{{n}} \:{an}\:{AP}} \\ $$$$ \\ $$$${a}_{\mathrm{15}} ={a}+\left(\mathrm{15}−\mathrm{1}\right){d}=\mathrm{32} \\ $$$${a}_{\mathrm{15}} ={a}+\mathrm{14}{d}=\mathrm{32}..................\left({i}\right) \\ $$$${S}_{\mathrm{9}} +{S}_{\mathrm{29}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{9}}{\mathrm{2}}\left(\mathrm{2}{a}+\mathrm{8}{d}\right)+\frac{\mathrm{29}}{\mathrm{2}}\left(\mathrm{2}{a}+\mathrm{28}{d}\right)=\mathrm{100} \\ $$$$\:\:\:\:\:\mathrm{9}\left({a}+\mathrm{4}{d}\right)+\mathrm{29}\left({a}+\mathrm{14}{d}\right)=\mathrm{100} \\ $$$$\:\:\:\:\:\mathrm{9}\left({a}+\mathrm{4}{d}\right)+\mathrm{29}\left(\mathrm{32}\right)=\mathrm{100} \\ $$$$\:\:\:\:\:{a}+\mathrm{4}{d}=\frac{\mathrm{100}−\mathrm{29}×\mathrm{32}}{\mathrm{9}}=−\mathrm{92}.....\left({ii}\right) \\ $$$$\:\left({i}\right)−\left({ii}\right):\:\:\:\:\mathrm{10}{d}=\mathrm{32}+\mathrm{92}=\mathrm{124} \\ $$$$\:\:\:\:\:\:\:\:\:{d}=\frac{\mathrm{124}}{\mathrm{10}}=\frac{\mathrm{62}}{\mathrm{5}} \\ $$$$\:\:\:\:\:\:\:\:\:{a}+\mathrm{4}\left(\frac{\mathrm{62}}{\mathrm{5}}\right)=−\mathrm{92}\Rightarrow{a}=−\mathrm{92}−\frac{\mathrm{248}}{\mathrm{5}}=−\frac{\mathrm{708}}{\mathrm{5}} \\ $$$${a}_{\mathrm{23}} ={a}+\mathrm{22}{d}=−\frac{\mathrm{708}}{\mathrm{5}}+\mathrm{22}\left(\frac{\mathrm{62}}{\mathrm{5}}\right)=\frac{\mathrm{656}}{\mathrm{5}} \\ $$$$\langle\bullet\rangle\boldsymbol{\mathrm{FORMULAS}}\:\boldsymbol{\mathrm{USED}}\langle\bullet\rangle \\ $$$$\begin{array}{|c|c|}{{a}_{{n}} ={a}+\left({n}−\mathrm{1}\right){d}}\\{{S}_{{n}} =\frac{{n}}{\mathrm{2}}\left[\:\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}\:\right]}\\\hline\end{array} \\ $$