Question

write the first four term of Ap,if a=-3 ,d=2

Answer 1

Given :

First term, a = - 3

Common Difference, d = 2

If we add the d ( Common Difference ) in the terms, we automatically get the next term.

Second Term, $\sf{a_2}$ = a + d = - 3 + 2 = - 1

Third Term, $\sf{a_3}$ = $\sf{a_2}$ + d = - 1 + 2 = 1

Fourth Term, $\sf{a_4}$ = $\sf{a_3}$ + d = 1 + 2 = 3

Hence,

AP formed —

$\bold{- 3, \ - 1, \ 1, \ 3, \ .....}$

Answer 2

Given that the first term of the AP is - 3 ( a ) and the common difference between the APs is 2 ( d ).



From the properties of arithmetic progressions, we know :

$a_{n}=a+(n-1)d$, where $a_{n}$ is the n th term, a is the first term of the AP, n is the number of AP and d is the common difference between the APs.

Now,
As we have to find out the value of first four terms of the APs, substituting the first four consecutive numbers in place of n.

Then,
• First term = a = - 3


• Second term = a + ( 2 - 1 )d

= > a + d
= > - 3 + 2
= > - 1
→ Second term of the AP is - 1 .


• Third term = a + ( 3 - 1 )d

= > - 3 + 2( 2 )
= > - 3 + 4
= > 1
→ Third term of the AP is 1.


• Fourth term = a + ( 4 - 1 )d

= > a + 3d
= > - 3 + 3( 2 )
= > - 3 + 6
= > 3
→ Fourth term of the AP is 3 .



Hence,
First four terms are - 3 , - 1 , 1 and 3.
$\:$