Question

the 5th term of an arithmetic sequence is 38 and the night the term is 66 what is its 25th term​

Answer 1

Answer

25th term of AP is 178

Given

The 5th term of an arithmetic sequence is 38 and the night the term is 66

To Find

25th term

Point to be noted

nth term of an AP ,

$\rm a_n=a+(n-1)d$

Solution

5th term of an AP is 38

$\to \rm a_5=38\\\\\to \rm a+(5-1)d=38\\\\\to \rm a+4d=38...(1)$

9th term of an AP is 66

$\to \rm a_9=66\\\\\to \rm a+(9-1)d=66\\\\\to \rm a+8d=66...(2)$

Solve eq. (1) - eq. (2) ,

⇒ (  a + 4d ) - ( a + 8d ) = 38 - 66

⇒ -4d = -28

⇒ d = 7

Sub. d value in (1) ,

⇒ a + 4(7) = 38

⇒ a + 28 = 38

⇒ a = 10

So , 25th term of an AP is ,

$\to \rm a_{25}=a+(25-1)d\\\\\to \rm a_{25}=10+24(7)\\\\\to \rm a_{25}=10+168\\\\\to \rm a_{25}=178$

Answer 2

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Given that,

a5 = 38

a9 = 66

We can write these terms as...

a5 → a + 4d = 38...... (1)

a9 → a + 8d = 66...... (2)

Now, subtract the equations (1) & (2), we get

➡ - 4d = - 28

➡ 4d = 28

➡ d = 7

Now, substitute the value of d in (1)

➡ a + 4(7) = 38

➡ a + 28 = 38

➡ a = 38 - 28

➡ a = 10

To find the value of 25th term,

↪ an = a + (n - 1)d

a = 10

d = 7

n = 25

➡ a25 = 10 + (25 - 1)(7)

➡ a25 = 10 + 24(7)

➡ a25 = 10 + 168

➡ a25 = 178