Question

the 5 term of an arithemetic sequence is 38 and 9th term is 66 fond the 25th term
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Answer 1

Given:

a5 = 38 and,

a9 = 66

We know, nth term is given as,

[an = a + (n-1)× d]

⇒ a5 = a + 4d ..(1)

And, a9 = a + 8d …(2)

Solving eq(1) and eq(2), we get,

a9 = a5 + 4d

⇒ 66 = 38 + 4d

⇒

.

Now, from eq (1),

a5 = a + 4d

⇒ 38 = a + 4(7)

⇒ a = (38-(4×7))

⇒ a = 38 – 28

⇒ a = 10

And,

25th term, a25 = a + 24d

= 10 + (24×7)

= 178

Hence, 25th term is 178

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