Question 2 (v)
Verify that each of the following is an AP and then write its next three terms.
a, 2a + 1, 3a + 2, 4a + 3, . . . .

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Solution

$H e r e, a_{1} = a, a_{2} = 2 a + 1, a_{3} = 3 a + 2 a n d a_{4} = 4 a + 3$
$a_{2} - a_{1} = 2 a + 1 - a = a + 1$
$a_{3} - a_{2} = 3 a + 2 - 2 a - 1 = a + 1$
$a_{4} - a_{3} = 4 a + 3 - 3 a - 2 = a + 1$
$a_{2} - a_{1} = a_{3} - a_{2} = a_{4} - a_{3} = a + 1 = C o m m o n d i f f e r e n c e$
Since the difference between successive terms are same, it forms an AP.
The next three terms are,
$a_{5} = a + 4 d = a + 4 (a + 1) = 5 a + 4$
$a_{6} = a + 5 d = a + 5 (a + 1) = 6 a + 5$
$a_{7} = a + 6 d = a + 6 (a + 1) = 7 a + 6$

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