Question

if the sum of the first n terms of an A.P is 4n-n^2.find the 1st,2nd,3rd and 10th and nth terms

Answer 1

Answer:

S1=t1=4×1-1^2

=3

S2=8-4=4

t2=4-3=1

c.d =1-3=-2

t3=1+(-2)

=-1

so

t10=3+(10-1)(-2)

=3-18

=-15

tn=3+(n-1)(-2)

=3-2n+2

=5-2n. Ans

Answer 2

Step-by-step explanation:

sum of the n terms = 4n - n^2

$s1 \: = 4(1) - {1}^{2} \\ = 4 - 1 \\ = 3 \\ s2 = 4(2) - {2}^{2} \\ = 8 - 4 \\ = 4 \\ s3 = 4(3) - {3}^{2} \\ = 12 - 9 \\ = 3$

a1 = s1 = 3

a2 = s2 - s1 = 4 - 3 = 1

a3 = s3 - s2 = 3 - 4 = -1

therefore the AP is

3, 1, -1.......

d = a2 - a1 = 1 - 3 = -2

10th term

a + 9d

= 3 + 9(-2)

= 3 - 18

= - 15

hope you get your answer