Answer:

Given: a  

n

​

=9−5n

Taking n=1,  

a  

1

​

=9−5(1)=9−5=4

Taking n=2,  

a  

2

​

=9−5(2)=9−10=−1

Taking n=3,  

a  

3

​

=9−5(3)=9−15=−6

Therefore the series is 4,−1,−6,...

So,a=4,d=a  

2

​

−a  

1

​

=−1−4=−5

Now, we have to find the sum of the first 15  

th

 terms of the AP

S  

n

​

=  

2

n

​

[2a+(n−1)d]

⇒S  

n

​

=  

2

15

​

[2×4+(15−1)(−5)]

⇒S  

15

​

=  

2

15

​

[8−70]

⇒S  

15

​

=  

2

15

​

[−62]

⇒S  

15

​

=15×(−31)

⇒S  

15

​

=−465

Hence, the sum of 15  

th

 terms is −465.

Step-by-step explanation:

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