Question

the 5th term of an as is 40 and 10th term is 20 find 15th term how many terms of this sequence make the sum zero

Answer 1

Answer:

15th term = 0

sum of 29 terms = 0

Step-by-step explanation:

5th term = a + 4d = 40

10th term = a + 9d = 20

10th - 5th

(a + 9d) - (a + 4d) = 20 - 40

a + 9d - a - 4d = -20

5d = - 20

d = -20/5 = -4

substitute d value in

a + 4d = 40

a + 4(-4) = 40

a - 16 = 40

a = 56

15th term is

a + 14d = 56 + 14(-4) = 56 - 56 = 0

$sum \: = 0 \\ \frac{n}{2} (2a + (n - 1)d) = 0 \\ \frac{n}{2} (2(56) + (n - 1) \times ( - 4)) \\ \frac{n}{2} (112 - 4n + 4) \\ \frac{n}{2} (116 - 4n) \\ \frac{n}{2} \times 2(58 - 2n) \\ 58n - 2 {n}^{2} = 0 \\ 2n(29 - n) = 0 \\ 2n \: = 0 \: \: \: \: 29 - n = 0 \\ we \: take \: 29 - n = 0 \\ n = 29 \\ therefore \: sum \: of \: 29 \: terms \: makes \: 0$

Answer 2

Solution,

Here, we have

a(5) = 40

a(10) = 20

To Find,

a(15) = ??

Here, we get

Here, we geta(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.