The AP whose first term is 10 and common difference is 3 is

10, 13, 16, 19, ...

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

5,7,9, 11, ...

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8, 12, 16,20, ...

No worries! We‘ve got your back. Try BYJU‘S free classes today!

All the above

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 10, 13, 16, 19, ...
If $a_{1}$ is the first term, then the nth term is $a_{1} + (n - 1) d$ , where $d$ is the common difference
Here $a_{1} = 10, d = 3,$
Hence,
$a_{2} = a_{1} + (2 - 1) d = 10 + 3 = 13$
$a_{3} = a_{1} + (3 - 1) d = 10 + 6 = 16$ and so on.....

Suggest Corrections

Similar questions

Q. Find the 12th term of the AP. whose first term is 9 and common difference is 10.

how many ap with 10 terms are there whose first term is in the set {1,2,3} and whose common difference is in the set {1,2,3,4,5}

The sum of the first n terms of an AP whose first term is 8 and the common difference is 20 is equal to the sum of the first 2n terms of another AP, whose first term is - 30 and the common difference is 8. find n?

Q. Question 33
The sum of the first n terms of an AP whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another AP whose first term is –30 and the common difference is 8. Find n.

Q. The sum of the first

n

terms of an

A P

whose first term is

8

and the common difference is

20

is equal to the sum of first

2 n

terms of another

A P

whose first term is

40

and the common difference is

6

. Find

n

Join BYJU'S Learning Program

The AP whose first term is 10 and common difference is 3 is

The correct option is C 10, 13, 16, 19, ...If a1 is the first term, then the nth term is a1+(n−1)d, where d is the common differenceHere a1=10,d=3,Hence,a2=a1+(2−1)d=10+3=13a3=a1+(3−1)d=10+6=16 and so on.....

The correct option is C 10, 13, 16, 19, ...
If $a_{1}$ is the first term, then the nth term is $a_{1} + (n - 1) d$ , where $d$ is the common difference
Here $a_{1} = 10, d = 3,$
Hence,
$a_{2} = a_{1} + (2 - 1) d = 10 + 3 = 13$
$a_{3} = a_{1} + (3 - 1) d = 10 + 6 = 16$ and so on.....