Arithmetic progression (A.P.)

 

 

Solution       Sn= 200    a= 20   d= -1

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

200=n/2[2(20)+(n-1)(-1)]

200=n/2[40-n+1]

200=n/2[41-n]

400=41n-n²    

n² – 41n+400=0

n² – 25n-16n+400=0

n(n-25)-16(n-25)=0

(n-25) (n-16)=0

n=25 and n=16

case1 when, n=25

a_n=a+(n-1)d

a_n=20+(25-1)(-1)

a_n=20+24(-1) = 20-24

a_n= -4  (not possible)

2^nd case (when n =16)

a_n=a+(n-1)d

a_n =20+(16-1)(-1)     

a_n= 20-15 = 5

number of rows are 16, and number  logs in top row are 5