write an A.P whose first term is a=6 common difference d=3
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Sequence of terms in A.P. = a , a + d , a + 2d , a + 3d.....a + ( n - 1 )d
{ where n is the last term }
In the question , a = 6 and common difference = - 3
→ first term = a = 6
→ second term = a + d
second term = 6 + ( - 3 )
second term = 6 - 3
second term = 3
→ third term = a + 2d
third term = 6 + 2( - 3 )
third term = 6 - 6
third term = 0
→ nth term ( last term ) = a + ( n - 1 )d
nth term = 6 + ( n - 1 )( - 3 )
nth term = 6 - 3n + 3
nth term = 9 - 3n
Therefore,
Required AP is 6 , 3 , 0 .....9 - 3( last term )
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