It the twelvth,85th and the last term of an arithmetic sequence are 38,257 and 395 respectively. Calculate how many terms are there in the sequence

131

Step-by-step explanation:

The n th term of an arithmetic sequence is

= a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Given a₁₂ = 38 and a₈₅ = 257 , then

a₁ +11d = 38 → (1)

a₁ + 84d = 257 → (2)

Subtract (1) from (2) term by term

73d = 219 ( divide both sides by 73 )

d = 3

Substitute d = 3 into (1)

a₁ + 11(3) = 38

a₁ + 33 = 38 ( subtract 33 from both sides )

a₁ = 5

Thus

5 + 3(n - 1) = 395 ( subtract 5 from both sides )

3(n - 1) = 390 ( divide both sides by 3 )

n - 1 = 130 ( add 1 to both sides )

n = 131

Thus there are 131 terms in the sequence

x = 25

step-by-step explanation:

a^2+b^2=c^2

7^2+24^2=c^2

49+576=625

sqrt625=25

x=25

b

step-by-step explanation: