Question

Find the sum of 12 terms of an A.P. whose nth term is given by an=4-3n​

Answer 1

Hope my answer helps you :-)

Step-by-step explanation:

a¹=3(1)+4=7

a²=3(2)+4=10

a³= 3(3)+4=13

d is 3

Sum of first 12 terms of an A. P .

Sum of first 12 terms of an A. P .is

Sum of first 12 terms of an A. P .is:: 12/2(7+7+(12-1)3)

Sum of first 12 terms of an A. P .is:: 12/2(7+7+(12-1)3):: 6(14+33)

Sum of first 12 terms of an A. P .is:: 12/2(7+7+(12-1)3):: 6(14+33):: 6*47

Sum of first 12 terms of an A. P .is:: 12/2(7+7+(12-1)3):: 6(14+33):: 6*47:: 282

Sum of first 12 terms of an A. P .is:: 12/2(7+7+(12-1)3):: 6(14+33):: 6*47:: 282Hence your answer is C

Sum of first 12 terms of an A. P .is:: 12/2(7+7+(12-1)3):: 6(14+33):: 6*47:: 282Hence your answer is C

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Answer 2

Answer:

-186

Step-by-step explanation:

1st term of AP = 4 - 3 (1) = 4 - 3 = 1

12th term of AP = 4 - 3 (12) = 4- 36 = -32

sum of 12 terms of A.P. = n/2 (a1 + a12 )

= 12/2 (1 -32) = 6× (-31) = -186