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Question-106457




Question Number 106457 by Rio Michael last updated on 05/Aug/20
Answered by bshahid010 last updated on 05/Aug/20
According to question   T_5 =50 & T_7 =25  Consider an AP with ist term a and   common difference d  then  T_5 =a+4d & T_7 =a+6d  a+4d=50 & a+6d=25  from above d=−((25)/2)  and a=100  Now T_(15) =a+14d=100−175=−75  and S_(20) =(n/2)(2a+(n−1)d)  S_(20) =((20)/2)(2×100−19(((25)/2)))
$${According}\:{to}\:{question}\: \\ $$$${T}_{\mathrm{5}} =\mathrm{50}\:\&\:{T}_{\mathrm{7}} =\mathrm{25} \\ $$$${Consider}\:{an}\:{AP}\:{with}\:{ist}\:{term}\:{a}\:{and}\: \\ $$$${common}\:{difference}\:{d} \\ $$$${then} \\ $$$${T}_{\mathrm{5}} ={a}+\mathrm{4}{d}\:\&\:{T}_{\mathrm{7}} ={a}+\mathrm{6}{d} \\ $$$${a}+\mathrm{4}{d}=\mathrm{50}\:\&\:{a}+\mathrm{6}{d}=\mathrm{25} \\ $$$${from}\:{above}\:{d}=−\frac{\mathrm{25}}{\mathrm{2}} \\ $$$${and}\:{a}=\mathrm{100} \\ $$$${Now}\:{T}_{\mathrm{15}} ={a}+\mathrm{14}{d}=\mathrm{100}−\mathrm{175}=−\mathrm{75} \\ $$$${and}\:{S}_{\mathrm{20}} =\frac{{n}}{\mathrm{2}}\left(\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}\right) \\ $$$${S}_{\mathrm{20}} =\frac{\mathrm{20}}{\mathrm{2}}\left(\mathrm{2}×\mathrm{100}−\mathrm{19}\left(\frac{\mathrm{25}}{\mathrm{2}}\right)\right) \\ $$

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