Question

The eighth term of an arithmatic sequence is 5 times third term, and the seventh term is nine greater than fourth term. Find the first five terms

Answer 1

Answer:

$\star\large\mathcal{\underline{\underline{PLEASE \: }}}\large\mathcal\red{\underline{\underline{F}}}\large\mathcal\green{\underline{\underline{OLL}}}\large\mathcal\pink{\underline{\underline{OW \: }}}\large\mathcal\blue{\underline{\underline{ME}}}\star$

Step-by-step explanation:

$given , \: a8 = 5×a3$

$and, \: a7 = 9+a4$

$=> a1 +7d= 5(a1+2d)$

$=> a1+7d=5a1+10d$

$=> -4a = 3d -----(1)$

$now, a7 = 9+a4$

$=> a1+6d= 9+a1+3d$

$=>a1-a1+ 6d-3d = 9$

$=> 3d = 9$

$=> d = \frac{9}{3} = 3$

$From \: (1) \: , we \: get:$

$-4a = 3(3)$

$= -4a = 9$

$= a = -\frac{9}{4}$

$now, \: a2 = a1+d$

$=> - \frac{9}{4} +3$

$=> \frac{ - 9 + 12}{4} = \frac{3}{4}$

$a3 = a2+d$

$=> \frac{3}{4} +3$

$=> \frac{3 + 12}{4} = \frac{15}{4}$

$a4 = a3+d$

$=> \frac{15}{4} +3$

$=> \frac{15 + 12}{4} = \frac{27}{4}$

$therefore, \: first \: five \: terms \: are :$

$- \frac{9}{4} \: , \: \frac{3}{4} , \: \frac{15}{4} , \: \frac{27}{4} , \: \frac{39}{4} .$

$please \: like \: my \: answer \\ \&\\ mark \: me \: brainliest \: please \: :)))$

$\star\large\mathcal{\underline{\underline{PLEASE \: }}}\large\mathcal\red{\underline{\underline{F}}}\large\mathcal\green{\underline{\underline{OLL}}}\large\mathcal\pink{\underline{\underline{OW \: }}}\large\mathcal\blue{\underline{\underline{ME}}}\star$

$\star \: i \: hope \: you \: will \: understand \: me\star$