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Question Number 206365 by hardmath last updated on 12/Apr/24

$$\mathrm{Number}\mathrm{series}:$$$${\mathrm{a}}_3=2\mathrm{a}+\mathrm{b}-6$$$${\mathrm{a}}_9=\mathrm{a}+\mathrm{b}+5$$$${\mathrm{a}}_{15}=3\mathrm{a}+\mathrm{b}-7$$$$\mathrm{Find}:\mathrm{a}=?$$

Commented by A5T last updated on 12/Apr/24

$$Arithmeticprogression?Is\mathrm{a}thefirstterm?$$

Commented by hardmath last updated on 12/Apr/24

$$\mathrm{yes},\mathrm{arithmetic}$$

Commented by A5T last updated on 12/Apr/24

$$Is\mathrm{a}thefirstterm?$$

Commented by hardmath last updated on 12/Apr/24

$$\mathrm{yes}\mathrm{dear}\mathrm{ser}$$

Answered by A5T last updated on 12/Apr/24

$$a_{15}-a_3=a-1=12d$$$$a_3-a_9=a-11=-6d$$$$\Rightarrow 18d=10\Rightarrow d=\frac{5}{9}\Rightarrow a=\frac{23}{3}$$$$a_3=a+2d=\frac{79}{9}=\frac{46}{3}-6+b\Rightarrow b=\frac{-5}{9}$$$$\Rightarrow a_3=\frac{79}{9},a_9=\frac{109}{9};a_{15}=\frac{139}{9}$$$$a_n=\frac{64+5n}{9}$$

Answered by Rasheed.Sindhi last updated on 13/Apr/24

$${\mathrm{a}}_{\mathrm{n}}\mathrm{is}\mathrm{an}\mathrm{AP}\left(given\right)$$$$\because 3,9,15arein\mathrm{AP}$$$$\therefore {\mathrm{a}}_3,{\mathrm{a}}_9,{\mathrm{a}}_{15}arealsoin\mathrm{AP}$$$$\therefore {\mathrm{a}}_9=\frac{{\mathrm{a}}_3+{\mathrm{a}}_{15}}{2}$$$$\mathrm{a}+\mathrm{b}+5=\frac{5\mathrm{a}+2\mathrm{b}-13}{2}$$$$5\mathrm{a}+2\mathrm{b}-13=2\mathrm{a}+2\mathrm{b}+10$$$$3\mathrm{a}=23$$$$\mathrm{a}=\frac{23}{3}$$

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