Question

write an A.P whose first term is a=6 common difference d=3​

Answer 1

$\huge\underline\mathrm\green{Answer}$

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 )