Test: nth Term of AP - Class 10 MCQ

We have a_n = a+(n-1)d
a₂₅ = a+(25-1)d=a+24d
a₁₂ = a+(12-1)d=a+11d
a₂₅ - a₁₂ = -52
a+24d-(a+11d)=-52
a+24d-a-11d=-52
13d= -52
d= -4

Highest two digit number divisible by 5 is 95
l=95, a=10 and d=5
l=a+(n-1)d
95=10+(n-1)5
(n-1)5=85
n-1=17
n=18

Common difference=¼-0=½-¼=¼
Next term =¾+¼=1
Next term =1+¼=5/4

We have an AP : 
Here a= 
l=a+(n-1)d

l=a+(n-1)d , d=109-114=-5
First negative term will be
0 > a+(n-1)d
0 > 114+(n-1)(-5)
0 > 114-5n+5
5n > 119
n > 119/5=23.8
So the first negative term will be 24th term.

The given AP is x – 7, x – 2, x + 3 …
So a= x-7, d=x - 2- x + 7 = x + 3 - x +2 = 5
l = a + (n - 1)d
=x - 7 + (15 - 1)5
=x - 7 + 70 = x + 63

The correct answer is b

  A=21

d=21

=420

n=?

=a+(n-1)d

420=21+(n-1)21

(n-1)*21=399

n-1=399/21

n-1=19

n=19+1=20

therefore the 20th term of the AP is 420

We have a=2, d=6-2=4
l=a+(n-1)d
210=2+(n-1)4
208/4=n-1
n=53rd term

The sequence is an arithmetic sequence

The nth term of an AP is given by:

APn = a + d(n - 1)

a = first term = 7

d = common ratio = 13 - 7 = 6 or 19 - 13 = 6 or 25 - 19 = 6

205 = 7 + 6(n - 1)

6(n - 1) = 198

n - 1 = 33

n = 34

The sequence has 34 terms.

We have an AP :10.0, 10.5, 11.0, 11.5
Where a=10, d=0.5
l=a+(n-1)d
 =10+(11-1)0.5
 =10+5=15