HirePro Sequence and Series Quiz-1

The 6th term is a + 5d = 6
The 16th term is a + 15d = 14
Equating these terms with their values and solving as simultaneous equations, we get a = 2 and d = 4/5. So the 27th term is (a + 26d) = 2 + (104/5) = 114/5.

The 2nd and the 7th terms are (a + d) and (a + 6d) respectively. The ratio of these terms is 1/3. Solving this ratio, we get 2a = 3d. The 5th term is (a + 4d) = 11. Substituting for a, we get a = 3 and d = 2. Therefore, the 15th term is (a + 14d) = 31.

The AP can be expressed as a, (a + d), ---, (a +12d). The sum of the first 3 terms is (3a + 3d) = 8 and that of the last 3 is (3a + 33d) = 16. Solving these equations, we get a = 13/5 and d = 4/15. The sum of the middle 3 terms is (3a + 18d) = 25.

The 7th and the 10th terms are (a + 6d) and (a + 9d) respectively. The ratio of these terms is -1. Solving this ratio, we get -2a = 15d. The 16th term is (a + 15d) = -15. Substituting for a, we get a = 15 and d = -2. Therefore, the 3rd term is (a + 2d) = 11.

Explanation:
Given, a=20, a12= 120, n=12.
Sn =n/2 (a+l)
=> S12 =12/2(20+120)
= 6*140 = 840.

Explanation:
Let a be the first term and r be the common ratio of the G. P., then ar3 = 8 …(1)
ar7=128 …(2)
Dividing 2 by 1:
r4 = 16
r = 2
putting r=2 in equation(1)
we get a=1.
So the required GP= 1,2,4,8,16…

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Solution:
Given AP: 28, 22, x, y, 4
Here, first term, a = 28
Common difference, d = 22 – 28 = -6
Hence, x = 22 – 6 = 16
y = 16 – 6 = 10.
Hence, the values of x and y are 16 and 10, respectively.

Solution:
Given AP: 1 , 5 , 9 , 13 , 17, 21, 25, …
Here, a = 1
d = 5-1 = 4
The formula to find the nth term of an AP is a + (n-1)d.
The nth term of an AP = 1 + (n-1)4
= 1 + 4n - 4
= 4n - 3
Hence, the nth term of AP 9, 13, 17, 21, 25, … is 4n-3.

Solution:
Given sequence is 4, 7, 10, 13, …
Here, a = 4
d = 7-4 = 3
The general term of an AP is:
an = a+(n-1)d
an = 4 +(n-1)3
an = 4 + 3n-3
an = 3n+1
Therefore, the general term of the series 4, 7, 10, 13 is 3n+1.

Solution:
Given AP: 27, 24, 21, …
Here, a = 27
d = 24 – 27 = -3.
Also given that an = 0
Now, we have to find the value of n.
Hence, an = a+(n-1)d
0 = 27 +(n-1)(-3)
0 = 27 -3n +3
0 = 30 – 3n
3n = 30
Hence, n = 10
Therefore, the 10th term of AP is 0.