Question

The 5 th term of an arithmetic sequence is 40 and 10th term 20. Find its
15th term. How many terms of this sequence makes the sum 0.
Pls ans fast

Answer 1

Step-by-step explanation:

a5= 40

a10=20

a+4d=40

a+9d= 20(sub)

-5d= 20

d= -4

sub d value in any one of the eq

a+4(-4)=40

a-16=40

a= 40+16

a= 56

a15= a+14d

= 56+14(-4)

= 56-56

= 0

Sn= n/2[2a+(n-1)d]

0 = n/2[ 112 +(n-1)(-4)]

0 = n/2[ 112-4n+4]

0 = n[ 116-4n]

116-4n= 0

116= 4n

n= 116/4

n= 29

29 terms of the sequence makes the sum 0

Answer 2

Answer:

Step-by-step explanation:

Solution,

Here, we have

a(5) = 40

a(10) = 20

To Find,

a(15) = ??

Here, we get

a(10) - a(5) = 5d

⇒ 5d = 20

⇒ d = 20/5

⇒ d = - 4

We know,

⇒ a(15) = a(10) + 5d

⇒ a(15) = 20 + 5 × (- 4)

⇒ a(15) = 20 - 20

⇒ a(15) = 0

Hence, the 15th term is 0.

Now, 2nd part of the question,

Let 1st term of A.P. be a.

Here, we get

a = 56

Now, we know that,

S(n) = n/2[2a + (n - 1)d]

On putting all the values, we get

S(n) = n/2[2a + (n - 1)d] = 0

⇒ S(n) = n/2[2 × 56 +(n - 1) × (- 4)] = 0

⇒ S(n) = n/2[116 - 4n] = 0

⇒ S(n) = n[116 - 4n] = 0

⇒ S(n) = 29.

Hence, 29 terms of this sequence makes the sum 0.