Question

In an arithmetic sequence,the 11th term is double of the 5th term it's 8th term is 27,find its 18th term ?

Answer 1

from the question $A_{11} =2times A_{5}$ ⇒

$a+(11-1d)=a+(5-1)d times2$ ⇒

$a+10d=2a+8d$ ⇒

$a-2a=8d-10d$ ⇒

$-a=-2d$ ⇒

$a=2d$

let the 8th term $a_{8} =a+(8-1)d=27$ be equation (1) ⇒

$27=a+7d$ ⇒

$27=2d+7d$ in substituting $a=2d$  in equation (1) ⇒

$27=9d$ ⇒

$d=3$

∴ the 18 term ⇒ $a_{18} =a+(18-1)d$

= $a+17d=2d+17(3)$ ⇒

$a_{18} =2d+51$ ⇒ $a_{18} =51+2d$ ⇒

$a_{18} =56$ and $a=5$ since $d=3$

Answer:

Answer 2

Answer:

ans is correct

check!!!