Question

The 5th term of an arithmetic sequence is 38 and the 9th term is 66.what is its 25th term

Answer 1

25th term is equal to 178.
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Answer 2

The 25th term of the A.P. is 178.

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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 5th term of an arithmetic sequence is 38, so we can form an equation as,

$a_5 = a + (5 - 1)d = 38$

$\implies a + 4d = 38$ . . . (1)

The 9th term of the arithmetic sequence is 66, so we can form an equation as,

$a_9 = a + (9 -1)d = 66$

$\implies a + 8d = 66$  . . . (2)

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

a + 8d = 66

a + 4d = 38

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4d = 28

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∴ d = 7

On substituting d = 7 in eq. (1), we get

$a + (4 \times 7) = 38$

⇒ $a + 28 = 38$

⇒ $a = 38 - 28$

⇒ $\bold{a = 10}$

Therefore,

The 25th term of the A.P. is,

= $a_2_5$

= $a + (25 - 1)d$

= $a + 24d$

on substituting a = 10 and d = 7, we get

= $10 + (24\times 7)$

= $10 + 168$

= $\bold{178}$

Thus, the 25th term of an A.P. is 178.

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