Question

if 228 logs are to be stacked in a store in the following manner : 30 logs in the bottom,28 in the next row,then 26 and so on . in how many rows can these 228 logs be stacked?how many logs are there in the last row ?

Answer 1

S n=228,

AP:30,28,26,........

a=30,d=-2

Sn=n/2[2a+(n-1)d]

228=n/2[60+(n-1)(-2)]

=n/2[60-2n+2]

=n/2[62-2n]

=n(31-n)

=-n^2+31n

n^2-31n+228=0

(n-19)(n-12)=0

n=19,12

last row=tn=a+(n-1)d

=30+(12-1)×(-2)

30-22=8 logs. when we put n=19, last row= -6 logs , which is not possible.

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