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Question Number 10583 by Saham last updated on 19/Feb/17

The sum of the 4^(th )  and 6^(th ) terms of an AP is 42. the sum of  the third and 9th terms of the proression is 52. Find the  first term , the common difference and the sum of the first  10 terms of the progression.

$$\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{4}^{\mathrm{th}\:} \:\mathrm{and}\:\mathrm{6}^{\mathrm{th}\:} \mathrm{terms}\:\mathrm{of}\:\mathrm{an}\:\mathrm{AP}\:\mathrm{is}\:\mathrm{42}.\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{third}\:\mathrm{and}\:\mathrm{9th}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{the}\:\mathrm{proression}\:\mathrm{is}\:\mathrm{52}.\:\mathrm{Find}\:\mathrm{the} \\ $$$$\mathrm{first}\:\mathrm{term}\:,\:\mathrm{the}\:\mathrm{common}\:\mathrm{difference}\:\mathrm{and}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first} \\ $$$$\mathrm{10}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{the}\:\mathrm{progression}. \\ $$

Commented by Saham last updated on 19/Feb/17

i have reduce it sir.

$$\mathrm{i}\:\mathrm{have}\:\mathrm{reduce}\:\mathrm{it}\:\mathrm{sir}. \\ $$

Commented by sandy_suhendra last updated on 19/Feb/17

are you Tawakalitu?

$$\mathrm{are}\:\mathrm{you}\:\mathrm{Tawakalitu}? \\ $$

Commented by Saham last updated on 20/Feb/17

yes sir. i used my surname now.

$$\mathrm{yes}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{used}\:\mathrm{my}\:\mathrm{surname}\:\mathrm{now}. \\ $$

Answered by ajfour last updated on 19/Feb/17

2a+8d=42  2a+10d=52  so d=5; a=1  T_1 +T_2 +...+T_(10) =5(2+45)=235.

$$\mathrm{2}{a}+\mathrm{8}{d}=\mathrm{42} \\ $$$$\mathrm{2}{a}+\mathrm{10}{d}=\mathrm{52} \\ $$$${so}\:{d}=\mathrm{5};\:{a}=\mathrm{1} \\ $$$${T}_{\mathrm{1}} +{T}_{\mathrm{2}} +...+{T}_{\mathrm{10}} =\mathrm{5}\left(\mathrm{2}+\mathrm{45}\right)=\mathrm{235}. \\ $$$$ \\ $$

Answered by mrW1 last updated on 19/Feb/17

a_3 =a+2d  a_4 =a+3d  a_6 =a+5d  a_9 =a+8d    a_4 +a_6 =42  ⇔(a+3d)+(a+5d)=42  2a+8d=42  a+4d=21   ...(i)    a_3 +a_9 =52  ⇔(a+2d)+(a+8d)=52  2a+10d=52  a+5d=26   ...(ii)    (ii)−(i): d=5   (common difference)  a=21−4×5=1   (first term)    a_(10) =a+9d=1+9×5=46    S_(10) =((10(1+46))/2)=235

$${a}_{\mathrm{3}} ={a}+\mathrm{2}{d} \\ $$$${a}_{\mathrm{4}} ={a}+\mathrm{3}{d} \\ $$$${a}_{\mathrm{6}} ={a}+\mathrm{5}{d} \\ $$$${a}_{\mathrm{9}} ={a}+\mathrm{8}{d} \\ $$$$ \\ $$$${a}_{\mathrm{4}} +{a}_{\mathrm{6}} =\mathrm{42} \\ $$$$\Leftrightarrow\left({a}+\mathrm{3}{d}\right)+\left({a}+\mathrm{5}{d}\right)=\mathrm{42} \\ $$$$\mathrm{2}{a}+\mathrm{8}{d}=\mathrm{42} \\ $$$${a}+\mathrm{4}{d}=\mathrm{21}\:\:\:...\left({i}\right) \\ $$$$ \\ $$$${a}_{\mathrm{3}} +{a}_{\mathrm{9}} =\mathrm{52} \\ $$$$\Leftrightarrow\left({a}+\mathrm{2}{d}\right)+\left({a}+\mathrm{8}{d}\right)=\mathrm{52} \\ $$$$\mathrm{2}{a}+\mathrm{10}{d}=\mathrm{52} \\ $$$${a}+\mathrm{5}{d}=\mathrm{26}\:\:\:...\left({ii}\right) \\ $$$$ \\ $$$$\left({ii}\right)−\left({i}\right):\:{d}=\mathrm{5}\:\:\:\left({common}\:{difference}\right) \\ $$$${a}=\mathrm{21}−\mathrm{4}×\mathrm{5}=\mathrm{1}\:\:\:\left({first}\:{term}\right) \\ $$$$ \\ $$$${a}_{\mathrm{10}} ={a}+\mathrm{9}{d}=\mathrm{1}+\mathrm{9}×\mathrm{5}=\mathrm{46} \\ $$$$ \\ $$$${S}_{\mathrm{10}} =\frac{\mathrm{10}\left(\mathrm{1}+\mathrm{46}\right)}{\mathrm{2}}=\mathrm{235} \\ $$

Commented by Saham last updated on 19/Feb/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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