Question

A teacher wrote a couple of terms on the blackboard. The terms are;
(a) 3n + 1
(b) 2n² + 3
(c) n³ + n
(d) 4n² + n

(i) Which of the following can be the nth term of an AP?
(ii) Write the 20th term of the AP answered by you in part (i).

Class 10, Math.
28 Points.

Answer 1

(a) 3n + 1 is the nth term of an AP.

(ii) The 20th term of the AP is

term of this AP are 4,7,10..........

a=4 d=3 and n= 20

a20 = a+(n-1)d

a20 = 4+(20-1)3

= 4+ 19(3)

= 4+57

= 61

hope you satisfied with my answer.

please mark as brainly.

Answer 2

Answer:

To find the nth term you must know about the Ap formed by these terms

that term would be the nth term whose A.P is correct

Taking term (a) 3n+1

put n=1. (to get the 1st term )

3n+1

3×1+1

3+1=4

a1=4

put n=2

3×2+1

6+1=7

a2= 7

put n=3

3×3+1

9+1=10

a3=10

common difference in a1,a2,a3 (4,7,10) = 3

now checking (b) 2n²+3

put n=1

2n²+3

2×1+3=5

a1=5

put n=2

2×2²+3

2×4+3.

8+3=11

a2=11

put n=3

2×3²+3

2×9+3=18+3=21

a3=21

common difference between a1,a2,a3,(5,11,21) is not equal ...i.e it is variable not constant so it is not common [11-5=6 and 21-11=10]

so these both equation are not equal that's why we conclude that 2n²+3 is not nth term

i told u the way to check the nth term u can check the other 2 nth term

The nth term is 3n+1

(ii) According to the question we have to find the 20 th term

put n=20 to get 20 th term

3×20+1

60+1

20 th term is 61

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