Verify that each of the following is an $ \mathrm{AP} $, and then write its next three terms.
$ a, 2 a+1,3 a+2,4 a+3, \ldots $
Given:
Given sequence is \( a, 2 a+1,3 a+2,4 a+3, \ldots \)
To do:
We have to verify whether the given sequence is an AP and write its next three terms.
Solution:
In the given sequence,
$a_1=a, a_2= 2a+1, a_3=3a+2, a_4=4a+3$
$a_2-a_1=(2a+1)-a=a+1$
$a_3-a_2=(3a+2)-(2a+1)=a+1$
$a_4-a_3=(4a+3)-(3a+2)=a+1$
Therefore,
$a_2-a_1=a_3-a_2=a_4-a_3$
The given sequence is an AP.
$d=a+1$
$a_5=a_4+d=(4a+3)+(a+1)=5a+4$
$a_6=a_5+d=(5a+4)+(a+1)=6a+5$
$a_7=a_6+d=(6a+5)+(a+1)=7a+6$
The next three terms of the given sequence are $(5a+4), (6a+5)$ and $(7a+6)$.
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