Question

find the first four terms of the sequence whose 'nth' term is 3n+1

Answer 1

Answer:

a=4

a+d=7

a+2d=10

a+3d=13

I hope it's helps you...

Answer 2

$\huge\mathfrak\red{Answer:-}$

We need to put different values of n here.

tn = 3n+1

put n=1

t(1)= 3×1 +1 = 4

n=2

t(2)= 3×2 +1 =7

n=3

t(3) = 3×3 +1 =10

n=4

t(4) = 3×4+1 = 12+1 = 13

so we can see that it makes an A.P with common difference 3

series is 4,7,10,13........

Now if we need to prove this then:

The given sequence is=

4,7,10,13.........(3n+1)

Let us understand :

When we put 1 in the place of n =

(3n+1)= 3×1 + 1= 4

Again puting 2 in the place of n

(3n+1) = 7

Similarly putting 3,4....in the place of 'n'

we get= 10,13......

Hence,the sequence is = 4,7,10,13........

Here 'a'= 4

C.d. = 3

Nth term= a+( n-1)d

Putting the value

= 4+ ( n-1)×3

= 4+ 3n -3

= 3n +4 -3

= ( 3n + 1 ) [Hence Proved]