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Let-T-k-is-the-k-th-term-and-S-k-is-the-sum-of-the-first-k-term-in-arithmetic-progression-If-T-3-T-6-T-9-T-12-T-15-T-18-45-Find-S-20-

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Question Number 24677 by Joel578 last updated on 24/Nov/17
Let T_k is the k^(th) term and S_k is the sum of the first k term in arithmetic progression If T_3 + T_6 + T_9 + T_(12) + T_(15) + T_(18) = 45 Find S_(20)
$$\mathrm{Let}\:{T}_{{k}} \:\mathrm{is}\:\mathrm{the}\:{k}^{\mathrm{th}} \:\mathrm{term}\:\mathrm{and}\:\mathrm{S}_{{k}} \:\mathrm{is}\:\mathrm{the}\:\mathrm{sum} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:{k}\:\mathrm{term}\:\mathrm{in}\:\mathrm{arithmetic}\:\mathrm{progression} \\ $$$$\mathrm{If}\:{T}_{\mathrm{3}} \:+\:{T}_{\mathrm{6}} \:+\:{T}_{\mathrm{9}} \:+\:{T}_{\mathrm{12}} \:+\:{T}_{\mathrm{15}} \:+\:{T}_{\mathrm{18}} \:=\:\mathrm{45} \\ $$$$\mathrm{Find}\:{S}_{\mathrm{20}} \\ $$
Answered by jota+ last updated on 24/Nov/17
6T_1 +(2+5+8+11+14+17)r=45 2T_1 +19r=15 T_1 +(T_1 +19r)=15 T_1 +T_(20) =15 S_(20) =15×10
$$\mathrm{6}{T}_{\mathrm{1}} +\left(\mathrm{2}+\mathrm{5}+\mathrm{8}+\mathrm{11}+\mathrm{14}+\mathrm{17}\right){r}=\mathrm{45} \\ $$$$\mathrm{2}{T}_{\mathrm{1}} +\mathrm{19}{r}=\mathrm{15} \\ $$$${T}_{\mathrm{1}} +\left({T}_{\mathrm{1}} +\mathrm{19}{r}\right)=\mathrm{15} \\ $$$${T}_{\mathrm{1}} +{T}_{\mathrm{20}} =\mathrm{15} \\ $$$${S}_{\mathrm{20}} =\mathrm{15}×\mathrm{10} \\ $$
Commented by Joel578 last updated on 22/Dec/17
thank you very much
$${thank}\:{you}\:{very}\:{much} \\ $$
Answered by ajfour last updated on 22/Dec/17
3(T_3 +T_6 +T_9 +...+T_(18) )=135 ⇒ (T_2 +T_3 +T_4 )+(T_5 +T_6 +T_7 )+.. ....+(T_(17) +T_(18) +T_(19) )=135 ..(i) and (((T_1 +T_(20) )/2))×20=S_(20) ⇒ T_1 +T_(20) =(S_(20) /(10)) ....(ii) adding (i) and (ii) S_(20) =135+(S_(20) /(10)) ⇒ S_(20) =135×((10)/9) = 150 .
$$\mathrm{3}\left({T}_{\mathrm{3}} +{T}_{\mathrm{6}} +{T}_{\mathrm{9}} +…+{T}_{\mathrm{18}} \right)=\mathrm{135} \\ $$$$\Rightarrow\:\:\left({T}_{\mathrm{2}} +{T}_{\mathrm{3}} +{T}_{\mathrm{4}} \right)+\left({T}_{\mathrm{5}} +{T}_{\mathrm{6}} +{T}_{\mathrm{7}} \right)+.. \\ $$$$\:\:\:\:\:….+\left({T}_{\mathrm{17}} +{T}_{\mathrm{18}} +{T}_{\mathrm{19}} \right)=\mathrm{135}\:\:\:\:..\left({i}\right) \\ $$$${and}\:\:\:\left(\frac{{T}_{\mathrm{1}} +{T}_{\mathrm{20}} }{\mathrm{2}}\right)×\mathrm{20}={S}_{\mathrm{20}} \\ $$$$\Rightarrow\:\:\:{T}_{\mathrm{1}} +{T}_{\mathrm{20}} =\frac{{S}_{\mathrm{20}} }{\mathrm{10}}\:\:\:\:\:\:\:….\left({ii}\right) \\ $$$${adding}\:\left({i}\right)\:{and}\:\left({ii}\right) \\ $$$${S}_{\mathrm{20}} =\mathrm{135}+\frac{{S}_{\mathrm{20}} }{\mathrm{10}} \\ $$$$\Rightarrow\:\:\:{S}_{\mathrm{20}} =\mathrm{135}×\frac{\mathrm{10}}{\mathrm{9}}\:=\:\mathrm{150}\:. \\ $$

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