CBSEi Class 10 Sequences (AP and GP) CORE

From (1) and (2), 3k 8 = 4 k
or, k = 3.
Thus, for k=3, the given three terms will form an AP.
Example 6: Find AP whose first term is —2 and common difference is —5.
Solution: Here a = 2, d = 5
So, the required AP will be
a, a+d, a+2d,……
or 2, 2+( 5), 2 + 2( 5),….
or, 2, 7, 12,…..
General Term of an AP
Recall that if a is the first term and d is the common difference, then the AP is
written as
a, a+d, a+2d, ….
Note that
first term( ) = a = a + (1 1) d
second term ( ) = a + d = a + (2 1) d
third term ( ) = a + 2d = a + (3 1) d
fourth term ( ) = a + 3d = a + (4 1) d
…….
tenth term ( ) = a + 9d = a + (10 – 1) d
50 th term ( ) = a + 49d = a + (50 – 1) d
Observing the pattern (coefficient of d in any term is 1 less than the term number),
we can write nth term ( )= a + (n 1) d
= a + (n 1) d
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