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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-t




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.
Thesumofthe4thand6thtermsofanAPis42.thesumofthethirdand9thtermsoftheproressionis52.Findthefirstterm,thecommondifferenceandthesumofthefirst10termsoftheprogression.
Commented by Saham last updated on 19/Feb/17
i have reduce it sir.
ihavereduceitsir.
Commented by sandy_suhendra last updated on 19/Feb/17
are you Tawakalitu?
areyouTawakalitu?
Commented by Saham last updated on 20/Feb/17
yes sir. i used my surname now.
yessir.iusedmysurnamenow.
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.
2a+8d=422a+10d=52sod=5;a=1T1+T2++T10=5(2+45)=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
a3=a+2da4=a+3da6=a+5da9=a+8da4+a6=42(a+3d)+(a+5d)=422a+8d=42a+4d=21(i)a3+a9=52(a+2d)+(a+8d)=522a+10d=52a+5d=26(ii)(ii)(i):d=5(commondifference)a=214×5=1(firstterm)a10=a+9d=1+9×5=46S10=10(1+46)2=235
Commented by Saham last updated on 19/Feb/17
God bless you sir.
Godblessyousir.

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