Question

consider an arithmetic sequence whose 7th term is 34 and 15 term is 66.find it's 25th term

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Answer 1

Step-by-step explanation:

let first term = a

common difference=d

ATQ

a+6d= 34

- (a+ 14d)= -66

-8d=34-66

d=32/8

d=4

putting d=4 in first eq.

a+6 d= 34

a+24=34

a=10

25 term =a+24d=10+96=106

Answer 2

$\huge\underline{\underline{\bf \bigstar ANSWER \bigstar}}$

Given ,

➤ 7th term = 34

➤ 15th term = 66

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 $\bf Common \ Difference = \dfrac{Term \ difference }{Position \ difference}$

   $\hookrightarrow \sf d = \dfrac{66-34}{15-7}$

          $\sf = \dfrac{32}{8}$

          $\bf = 4$

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 ➣ 25th term = 15th term + 10d

                     = 66 + 10 × 4

                     = 66 + 40

                     = 106

HOPE IT HELPS U ........... :)