Question

If the sum of the first n terms of an Ap is 4n-n^2,what is the first term?​

Answer 1

Answer:

hey here is your answer

According to question s= 4n-n*n or s= 4n-n2 s1 =4*1=1*1 = 4 – 1 = 3 first term=3 Now, sum of first two terms = s2 =4*2−2 *2 =8−4=4 Therefore Second term =s 2 −s 1 =4−3=1 =s3 =4×3−3 2 = 12 – 9 = 3 Therefore Third term = s 3 −s 2 = 3 – 4 = – 1 s 9 =4×9−9 2 = 36 – 81 = – 45 and, s 10 =4×10−10 2 = 40 – 100 = – 60 Therefore Tenth term = s 10 −s 9 = – 60 – (– 45) = – 60 + 45 = – 15 Also, s n =4n−n 2 and s n−1 =4(n−1)−(n−1) 2 =4n−4−n 2 +2n−1 =−n 2 +6n−5 Therefore, nth term = s n −s n−1 =4n−n 2 −(−n 2 +6n−5) =4n−n 2 +n 2 −6n+5=5−2n

Sn = 4n - n2

Put n = 1, we get

S1 = 4*1 - 12

= 4 – 1

= 3

So first term = 3

Now, sum of first two terms S2 = 4*2−22 (Put n=2)

= 8−4

= 4

So sum of first two terms = 4

Therefore Second term =S2 −S1

=4−3

=1

So second term = 1

Again S3 = 4×3 - 32 (Put n= 3)

= 12 – 9

= 3

Therefore Third term = S3 − S2

= 3 – 4

= – 1

So third term = -1

Again

S9 = 4×9−92 (Put n = 9)

= 36 – 81

= – 45

and

S10 = 4×10−102 (Put n = 10)

= 40 – 100

= – 60

Therefore Tenth term = S10 − S9

= – 60 – (– 45)

= – 60 + 45

= – 15

Now

Sn = 4n−n2

and Sn-1 = 4(n−1)−(n−1)2

= 4n − 4 − (n2 - 2n + 1)

= 4n − 4 − n2 + 2n - 1

= −n2 + 6n - 5

Therefore, nth term = Sn − Sn-1

= 4n−n2 −(−n2 +6n−5)

= 4n − n2 + n2 − 6n + 5

= 5 − 2n

So nth term of AP is 5 − 2n

hope it help you ☺️

Answer 2

Answer:

this is your require answer