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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 ])))

a15=32,S9+S29=100,a23=?AssumingananAP―a15=a+(15−1)d=32a15=a+14d=32..................(i)S9+S29=92(2a+8d)+292(2a+28d)=1009(a+4d)+29(a+14d)=1009(a+4d)+29(32)=100a+4d=100−29×329=−92.....(ii)(i)−(ii):10d=32+92=124d=12410=625a+4(625)=−92⇒a=−92−2485=−7085a23=a+22d=−7085+22(625)=6565⟨∙⟩FORMULASUSED⟨∙⟩an=a+(n−1)dSn=n2[2a+(n−1)d]

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