Question

The sum of the first 12 terms of an AP is 222 and the sum of the first 5 terms is 40. Write out the of the first four terms of the series

     

Answer 1

$S_{12}=222, and , S_5=40$

$S_n=n/2(a+(n-1)d)$

at $n=12$

$S_{12}=12/2(a+(12-1)d)=6(a+11d)=222$

$a+11d=37 .........(i)$

at $n=5, S_5=5/2(a+4d)=40$

$a+4d=(40\times2)/5$

$a+4d=16.........(ii)$

from (i)

$37-11d+4d=16$

$37-7d=16$

$7d=37-16=21$

$d=3$

from (i)

$a=37-11d=37-11\times3=4$

hence the first four terms are;

$a, a+d, a+2d, a+3d$

$4, 4+3, 4+2\times 3, 4+3\times 3$

$4, 7, 10, 13$