Problem 1 :
In a theater, there are 20 seats in the front row and 30 rows were allotted. Each successive row contains two additional seats than its front row. How many seats are there in the last row?
Solution :
Number of seats in 1^{st} row = 20
Number of seats in 2^{nd} row = 20 + 2 = 22
Number of seats in 3^{rd} row = 22 + 2 = 24
Total number of rows in theatre (n) = 30
a = 20, d = 2 and n = 30
number of seats in last row be "tn" or "l"
n = [(l - a)/d] + 1
30 = [(l - 20)/2] + 1
(30 - 1) = (l - 20)/2
29 (2) = l - 20
l - 20 = 58
l = 58 + 20
l = 78
So, the number of seats in the last row is 78.
Problem 2 :
The sum of three consecutive terms that are in A.P. is 27 and their product is 288. Find the three terms.
Solution :
Let the three terms are a - d, a and a + d.
Sum of three consecutive terms = 27
a - d + a + a + d = 27
3a = 27
a = 27/3 = 9
Product of three terms = 288
(a - d) a (a + d) = 288
a(a^{2} - d^{2}) = 288
9(9^{2} - d^{2}) = 288
(9^{2} - d^{2}) = 288/9
(9^{2} - d^{2}) = 32
- d^{2} = 32 - 81
- d^{2} = - 49
d = 7
1^{st} term = a - d = 9 - 7 = 2
2^{nd} term = a = 9
3^{rd} term = a + d = 9 + 7 = 16
So, the first three terms are 2, 9, 16.
Problem 3 :
The ratio of 6th and 8th term of an A.P. is 7:9. Find the ratio of 9th term to 13th term.
Solution :
t_{6} : t_{8} = 7 : 9
t_{6} / t_{8} = 7 / 9
(a + 5d) / (a + 7d) = 7/9
9(a + 5d) = 7(a + 7d)
9a + 45d = 7a + 49d
9a - 7a + 45d - 49d = 0
2a - 4d = 0
a - 2d = 0
a = 2d
ratio of 9th term to 13^{th} term
t_{9} : t_{13} = (a + 8d) / (a + 12d)
= (2d + 8d) / (2d + 12d)
= 10d / 14d
= 5 / 7
= 5 : 7
So, the ratio of 9th term to 13^{th} term is 5 : 7.
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