Question

The 10th term of an arithmetic sequence is 20 and the 20th term is 10 ..Find 30th term?

Answer 1

Step-by-step explanation:

t10= 20

a+9d = 20

t20= 10

a+ 19d= 30

10d= 10

d= 1

a=11

now a+29d= 11+29= 40

Answer 2

The 30th term of the A.P. is 0.

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Let's understand a few concepts:

To calculate the nth term of an A.P. we will use the following formula:

$\boxed{\bold{a_n = a + (n - 1)d}}$

where aₙ = last term, a = first term, d = common difference between the terms and n = no. of terms

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Let's solve the given problem:

The 10th term of an arithmetic sequence is 20, so we can form an equation as,

$a _1_0 = a + (10 - 1)d = 20$

$\implies a + 9d = 20$ . . . (1)

The 20th term of the arithmetic sequence is 10, so we can form an equation as,

$a _2_0 = a + (20 - 1)d = 10$

$\implies a + 19d = 10$ . . . (2)

On subtracting both the equations (1) and (2), we get

a + 9d = 20

a + 19d = 10

-   -          -

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  - 10d = 10

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∴ d = -1

On substituting d = -1 in eq. (1), we get

$a + (9 \times -1) = 20$

⇒ $a - 9 = 20$

⇒ $a = 20 + 9$

⇒ $\bold{a = 29}$

Therefore,

The 30th term of the A.P. is,

= $a_3_0$

= $a + (30 - 1)d$

on substituting a = 29 and d = -1, we get

= $29 + (29 \times -1)$

= $29 - 29$

= $\bold{0}$

Thus, the 30th term of an A.P. is zero.

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