Question

.Write an A.P. whose first term is a and common difference is d. a= -1.25, d= 3.​

Answer 1

Answer:

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Step-by-step explanation:

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 )

Answer 2

Step-by-step explanation:

a= -1.25 , d=3

First term = a= -1.25

Second term = d = 3

Third term = a + 2d = -1.25 + 6 = 4.75

and so on

Hence, the required A.P. is 4.75, 4.75, 4.75 .............