Question

The general term of a sequence is give by $a_{n}=-4n+15$. Is the sequence an A.P.? If so, find its 15th term and the common difference.

Answer 1

Answer:

The sequence is an A.P,  it's common Difference is - 4 & it's 15th term  is - 45.  

Step-by-step explanation:

Given :  

an = - 4n + 15 ……………..(1)

On putting n = 1 in eq 1,  

a1= - 4(1) + 15  

a1 = - 4 + 15

a1 = 11

On putting n = 2 in eq 1,  

a2 = - 4(2) + 15  

a2 = - 8 + 15  

a2 = 7

On putting n = 3 in eq 1,  

a3= - 4(3) + 15  

a3 = - 12 + 15

a3 = 3

 

Common difference, d1 = a2 – a1  

d1 = 7 – 11

d1 = - 4

Common difference, d2 = a3 – a2  

d2 = 3 - 7

d2 = - 4  

Since, common Difference are equal i.e d1 = d2 = - 4

Therefore, A.P is in the sequence.

On putting n = 15 in eq 1,  

a15= - 4(15) + 15  

a15 = - 60 + 15

a15 = - 45

15th term of an A.P is - 45.  

Hence, the sequence is an A.P,  it's common Difference is - 4 & it's 15th term  is - 45.  

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Answer 2

the sequence whose general term is given by -4n+15 is an A.P. with common difference -4.

first term 11.

and 15th term is -45.