Question

The question is in the picture.

Answer 1

Part A:

Let's denote no of seats in first row with r1 , second row with r2.....and so on.

r1=5

Since next row will have 10 additional row each time when we move to next row,

So,

r2=5+10=15

r3=15+10=25

Using the terms r1,r2 and r3 , we can find explicit formula

r1=5=5+0=5+0×10=5+(1-1)×10

r2=15=5+10=5+(2-1)×10

r3=25=5+20=5+(3-1)×10

So for nth row,

rn=5+(n-1)×10

Since 5=r1 and 10=common difference (d)

rn=r1+(n-1)d

Since 'a' is a convention term for 1st term,

⇒rn=a+(n-1)d

which is an explicit formula to find no of seats in any given row.

Part B:

Using above explicit formula, we can calculate no of seats in 7th row,

r7=5+(7-1)×10

r7=5+(7-1)×10 =5+6×10

r7=5+(7-1)×10 =5+6×10 =65

which is the no of seats in 7th row.