Long method:
Let the first number be xx. We know that, consecutive odd numbers have difference of 22
So, the numbers are
color(blue)(x,x+2,x+4,x+6,x+8,x+10,x+12,x+14,x+16x,x+2,x+4,x+6,x+8,x+10,x+12,x+14,x+16
It is stated that the third term (x+4x+4) is equal to 99
rarrx+4=9→x+4=9
color(blue)(rarrx=9-4=5→x=9−4=5
So, insert the value of xx in the sequence to get the sum
rarr(5)+(5+2)+(5+4)+(5+6)+(5+8)+(5+8)+(5+10)+(5+12)+(5+14)+(5+16)→(5)+(5+2)+(5+4)+(5+6)+(5+8)+(5+8)+(5+10)+(5+12)+(5+14)+(5+16)
rarr5+7+9+11+13+15+17+19+21→5+7+9+11+13+15+17+19+21
color(green)(rArr108⇒108
Hope this helps!!!
