Exercise 5.2
Question 16:
Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12.
Given
a3 = 16
a + (3 − 1) d = 16
a + 2d = 16 ………………………(1)
AtQ,
a7 − a5 = 12
[a+ (7 − 1) d] − [a + (5 − 1) d]= 12
(a + 6d) − (a + 4d) = 12
2d = 12
d = 6
From equation (1),
a + 2 (6) = 16
a + 12 = 16
a = 4
Therefore, A.P. will be 4, 10, 16, 22, …
Question 17:
Find the 20th term from the last term of the A.P. 3, 8, 13, …, 253
Given A.P. is
3, 8, 13, …, 253
Common difference. is 5.
Reversing the term of this A.P
253, 248, 243, …, 13, 8, 5
For this A.P.,
a = 253
d = 248 − 253 = −5
n = 20
a20 = a + (20 − 1) d
a20 = 253 + (19) (−5)
a20 = 253 − 95
a = 158
20th term from the last term is 158.
Question 18:
The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the A.P.
In A.P. nth term is given by
an = a + (n − 1) d
a4 = a + (4 − 1) d
a4 = a + 3d
Similarly,
a8 = a + 7d
a6 = a + 5d
a10 = a + 9d
it is given, a4 + a8 = 24
a + 3d + a + 7d = 24
2a + 10d = 24
a + 5d = 12 ..............(1)
a6 + a10 = 44
a + 5d + a + 9d = 44
2a + 14d = 44
a + 7d = 22 .............(2)
On subtracting equation (1) from (2),
2d = 22 − 12
2d = 10
d = 5
From equation (1),
a + 5d = 12
a + 5 (5) = 12
a + 25 = 12
a = −13
a2 = a + d = − 13 + 5 = −8
a3 = a2 + d = − 8 + 5 = −3
Therefore, the first three terms of this A.P. are −13, −8, and −3.