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