The seats in an arena are arranged so that each row is longer than the row in front of it by the same number of seats. The 13th row has 71 seats and the 19th row has 95 seats. Determine the total number of seats in the front three rows combined.

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Solution

Since each raw is longer than same number of seats, the number of seats are in AP(arithmetic progression)
nth term of an AP is given by
a+(n-1)d
13th term is 71,that is
a+12d=71 eqn (1)
19th term is 95,that is
a+18d=95 eqn(2)
Subtract eqn (2) from (1)
We get, 6d=24
That is d=4
Substitute value of d in eqn (1)
a+48=71
a=71-48=23
That is 1st raw has 23 seats
2nd raw=a+d=23+4=27seats
3rd raw=a+2d=23+8=31seats
Therefore, total no. of seats in first three rawa=23+27+31=81 seats

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