Math, asked by ananthurukkmini9343, 1 year ago

The 10th term of an A.P. is 31 and 20th term is 71. Find the 30th term.
20. Find the value of 'p' for which one root of the quadratic equation px? - 14x + 8 =0 is 6 times
the other​

Answers

Answered by Swarup1998
1

1. The 10th term of an A.P. is 31 and 20th term is 71. Find the 30th term.

Solution: Let the first term of the A.P. be a and the common difference be d.

Then the 10th term is (a + 9d)

and the 20th term is (a + 19d).

Given,

a + 9d = 31 ..... (1)

a + 19d = 71 ..... (2)

Using (2) - (1), we get

10d = 40

i.e., d = 4

Then a = 31 - 9 (4)

i.e., a = 31 - 36

i.e., a = - 5

Hence the 30th term is (a + 29d)

= - 5 + 29 (4)

= - 5 + 116

= 111

2. Find the value of 'p' for which one root of the quadratic equation px² - 14x + 8 =0 is 6 times

the other.

Solution: The given equation is

px² - 14x + 8 = 0 ..... (1)

Let the two roots be k and 6k

Since k, 6k are roots of (1),

k + 6k = - (- 14)/p

or, 7k = 14/p ..... (2)

& k . 6k = 8/p

or, 6k² = 8/p ..... (3)

Dividing (3) by (2), we get

6k² / 7k = 8/p / 14/p

or, 6k/7 = 4/7

or, k = 2/3

From (2), we write

7 (2/3) = 14/p

or, 21/3 = 14/p

or, p = 14 * 3/21

or, p = 2

Therefore the value of p is 2.

Answered by santy2
0

Answer:

1) The 30th term of the AP is 111

2) The value of p in the quadratic equation is 3

Step-by-step explanation:

1) The nth term of an AP is given by:

aₙ = a₁ + (n - 1)d

d = common difference

a₁ = the first term

n = Number of terms

10th term:

31 = a₁ + (10 - 1)d

a₁ + 9d = 31

20th term

71 = a₁ + (20 - 1)d

a₁ + 19d = 71

We solve the equations simultaneously to get the value of d and a₁

a₁ + 9d = 31.................1)

a₁ + 19d = 71...............ii)

Subtracting ii from i we have:

10d = 40

d = 4

a₁ = 31 - 9 × 4

a₁ = -5

The 30th term is given by:

a₃₀ = -5 + (30 - 1)4

= -5 + 116 = 111

The 30th term is 111

2) px² - 14x + 8 = 0

We have that:

We have that:

a = p, b = -14, c = 8

Let one root of the quadratic equation be k, the other root will be 6k

The sum of the roots of the quadratic equation is given by:

= -b/a = -(-14/p) = 14/p

k + 6k = 14/p

7k = 14/p

p = 14/7k

p= 2/k

The products of the roots is given by:

= c/a

k × 6k = 8/p

6k² = 8/p

p = 8/6k²

p = 4/3k²

Equating the two values of p we have:

4/3k² = 2/k

4k = 6k²

k = 4/6 = 2/3

p = 2/k

p = 2 × 3/2 = 3

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