Unable to understand solution to AP/GP problem

Hi Mr Koh,

                  I encountered this particular question in my revision package:

A solution was provided for it:

However, I don't understand the part which I have underlined-why do we need to subtract 145 (T10)?

 Student  X         

Perhaps a slightly different manner of looking at things might help:

Change in height of the tree at the end of the 10th year is 145cm.

Height of the tree at the end of the 10th year is 775cm.

Then height of the tree at the end of the 11th year is 775+145r 

(r denotes the common ratio of the GP) and height of the tree at the end of the 12th year is 775+145r +145r^2, so on and so forth.

Continuing things in this manner,

At the end of the infinite year of growth, height of tree

= 775+(145r + 145r^2 +145r^3+.................) -----------------------(1)            

Noting that the terms in brackets for (1) belongs to the sum of infinity of a GP,            

(1) becomes 775+145r/(1-r) which is also = 2000

Migrating the quantity 775 to the RHS of the above equation gives

145r/(1-r) = 1225

Therefore 145r = 1225 -1225r =====> r= 245/274 (shown)

Hope this helps. Peace.

Best Regards,

Mr Koh