In an arithemetic sequence 20th term is 40 and common difference 5 find 5th and first term About the author Clara

Answer: (a) x 5 =40,x 10 =20 x 10 −x 5 =5d 5d=20−40=−20 ⇒d=−4 Now, x 15 =x 10 +5d=20−20=0 (b) First term, f=x 5 −4d=40−4(−4)=56 S n = 2 n[2f+(n−1)d] ⇒0= 2 n[2×56+(n−1)(−4)] ⇒n(112−4n+4)=0 ⇒n(116−4n)=0 ⇒=0,n=29 Thus, 29 terms of this sequence make the sum zero. Reply

Answer:(a) x

5

=40,x

10

=20

x

10

−x

5

=5d

5d=20−40=−20

⇒d=−4

Now, x

15

=x

10

+5d=20−20=0

(b) First term, f=x

5

−4d=40−4(−4)=56

S

n

=

2

n[2f+(n−1)d]

⇒0=

2

n[2×56+(n−1)(−4)]

⇒n(112−4n+4)=0

⇒n(116−4n)=0

⇒=0,n=29

Thus, 29 terms of this sequence make the sum zero.