Question
Is the given sequence $a, 2a, 3a, 4a,...$ forms an$ AP$? If it forms an $AP$, then find the common difference d and write the next three terms.
✓
Answer
from the given sequence, we can have
$a_{2}-a_{1}=2 a-a=a$
$a_{3}-a_{2}=3 a-2 a=a$
$ a_{4}-a_{3}=4 a-3 a=a$
since $a_{k+1}-a_{k}$ i.e. the common difference is the same for all values of k
Hence, the given sequence forms an AP.
Now the next three terms are:
$a_5= a + 4d = a + 4a = 5a$
$a_6= a + 5d = a + 5a = 6a$
$a_7= a + 6d = a + 6a = 7a$
Next three terms are: 5a, 6a and 7a
$a_{2}-a_{1}=2 a-a=a$
$a_{3}-a_{2}=3 a-2 a=a$
$ a_{4}-a_{3}=4 a-3 a=a$
since $a_{k+1}-a_{k}$ i.e. the common difference is the same for all values of k
Hence, the given sequence forms an AP.
Now the next three terms are:
$a_5= a + 4d = a + 4a = 5a$
$a_6= a + 5d = a + 5a = 6a$
$a_7= a + 6d = a + 6a = 7a$
Next three terms are: 5a, 6a and 7a
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