Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?
Given A.P. is : 3, 15,27,39.....
Here a1 = 3, a2 = 15
a3 = 27, a4 = 39
d = a2 – a1 = 15 – 3 = 12
a54 = a (54 – 1)d
= 3 + 53 x 12
= 3 + 636 = 639
Let ntn term be 132 more than a54 We know that,
an = a + (n - 1)d
a54 + 132 = a + (n - 1)d
639 + 132 = 3 + (n - 1) x 12
771 - 3 = 12(n - 1)
768 = 12(n - 1)
Write first four terms of the AP, when the first term a and the common difference d are given as follows:
a – 10, d = 10
Write first four terms of the AP, when the first term a and the common difference d are given as follows:
a = –2, d = 0
Write first four terms of the AP, when the first term a and the common difference d are given as follows:
a = 4, d = – 3
Write first four terms of the AP, when the first term a and the common difference d are given as follows:
a = – 1.25, d = – 0.25
For the following APs, write the first term and the common difference
3, 1, – 1, – 3, . . .
For the following APs, write the first term and the common difference
– 5, – 1, 3, 7, . . .
For the following APs, write the first term and the common difference
0.6, 1.7, 2.8, 3.9, . . .
Mock Test Series
