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Number-series-a-3-2a-b-6-a-9-a-b-5-a-15-3a-b-7-Find-a-

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Question Number 206365 by hardmath last updated on 12/Apr/24
Number series: a_3 = 2a + b − 6 a_9 = a + b + 5 a_(15) = 3a + b − 7 Find: a = ?
$$\mathrm{Number}\:\mathrm{series}: \\ $$$$\mathrm{a}_{\mathrm{3}} \:=\:\mathrm{2a}\:+\:\mathrm{b}\:−\:\mathrm{6} \\ $$$$\mathrm{a}_{\mathrm{9}} \:=\:\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{5} \\ $$$$\mathrm{a}_{\mathrm{15}} \:=\:\mathrm{3a}\:+\:\mathrm{b}\:−\:\mathrm{7} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{a}\:=\:? \\ $$
Commented by A5T last updated on 12/Apr/24
Arithmetic progression? Is a the first term?
$${Arithmetic}\:{progression}?\:{Is}\:\mathrm{a}\:{the}\:{first}\:{term}? \\ $$
Commented by hardmath last updated on 12/Apr/24
yes, arithmetic
$$\mathrm{yes},\:\mathrm{arithmetic} \\ $$
Commented by A5T last updated on 12/Apr/24
Is a the first term?
$${Is}\:\mathrm{a}\:{the}\:{first}\:{term}? \\ $$
Commented by hardmath last updated on 12/Apr/24
yes dear ser
$$\mathrm{yes}\:\mathrm{dear}\:\mathrm{ser} \\ $$
Answered by A5T last updated on 12/Apr/24
a_(15) −a_3 =a−1=12d a_3 −a_9 =a−11=−6d ⇒18d=10⇒d=(5/9)⇒a=((23)/3) a_3 =a+2d=((79)/9)=((46)/3)−6+b⇒b=((−5)/9) ⇒a_3 =((79)/9),a_9 =((109)/9);a_(15) =((139)/9) a_n =((64+5n)/9)
$${a}_{\mathrm{15}} −{a}_{\mathrm{3}} ={a}−\mathrm{1}=\mathrm{12}{d} \\ $$$${a}_{\mathrm{3}} −{a}_{\mathrm{9}} ={a}−\mathrm{11}=−\mathrm{6}{d} \\ $$$$\Rightarrow\mathrm{18}{d}=\mathrm{10}\Rightarrow{d}=\frac{\mathrm{5}}{\mathrm{9}}\Rightarrow{a}=\frac{\mathrm{23}}{\mathrm{3}} \\ $$$${a}_{\mathrm{3}} ={a}+\mathrm{2}{d}=\frac{\mathrm{79}}{\mathrm{9}}=\frac{\mathrm{46}}{\mathrm{3}}−\mathrm{6}+{b}\Rightarrow{b}=\frac{−\mathrm{5}}{\mathrm{9}} \\ $$$$\Rightarrow{a}_{\mathrm{3}} =\frac{\mathrm{79}}{\mathrm{9}},{a}_{\mathrm{9}} =\frac{\mathrm{109}}{\mathrm{9}};{a}_{\mathrm{15}} =\frac{\mathrm{139}}{\mathrm{9}} \\ $$$${a}_{{n}} =\frac{\mathrm{64}+\mathrm{5}{n}}{\mathrm{9}} \\ $$
Answered by Rasheed.Sindhi last updated on 13/Apr/24
a_n is an AP (given) ∵ 3,9,15 are inAP ∴ a_3 ,a_9 ,a_(15 ) are also in AP ∴ a_9 =((a_3 +a_(15) )/2) a + b + 5=((5a+2b−13)/2) 5a+2b−13=2a+2b+10 3a=23 a=((23)/3)
$$\mathrm{a}_{\mathrm{n}} \:\mathrm{is}\:\mathrm{an}\:\mathrm{AP}\:\left({given}\right) \\ $$$$\because\:\:\mathrm{3},\mathrm{9},\mathrm{15}\:{are}\:{in}\mathrm{AP} \\ $$$$\therefore\:\mathrm{a}_{\mathrm{3}} ,\mathrm{a}_{\mathrm{9}} ,\mathrm{a}_{\mathrm{15}\:} {are}\:{also}\:{in}\:\mathrm{AP} \\ $$$$\therefore\:\mathrm{a}_{\mathrm{9}} =\frac{\mathrm{a}_{\mathrm{3}} +\mathrm{a}_{\mathrm{15}} }{\mathrm{2}} \\ $$$$\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{5}=\frac{\mathrm{5a}+\mathrm{2b}−\mathrm{13}}{\mathrm{2}} \\ $$$$\mathrm{5a}+\mathrm{2b}−\mathrm{13}=\mathrm{2a}+\mathrm{2b}+\mathrm{10} \\ $$$$\mathrm{3a}=\mathrm{23} \\ $$$$\mathrm{a}=\frac{\mathrm{23}}{\mathrm{3}} \\ $$

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