Respuesta :

geerky42
a5 = 6a1 = a4 + d = a3 + 2d = ... = a1 + 4d

So we have 6a1 = a1 + 4d -> a1 = 4/5 d

Now for second part, we have sum of first six term = a1 + (a1+d) + a1+2d + ... + (a1 + 5d) = 6a1 + 15d

So 99 = 6a1 + 15d

Now we have two equations:
a1 = 4/5 d
99 = 6a1 + 15d

Solve the system of linear equation and we will get a1 = 4 and d = 5

Therefore the first term is 4 and common difference is 5.


briannagrant32

Answer:

Isn't the definition of 5th term (a + 4d)

and that of the first term a ?

so a+4d = 6a

4d = 5a

d = 5a/4 -----> **

Also : sum(6) = (6/2)(2a + 5d)

99 = 3(2a + 5(5a/4) )

33 = 2a + 25a/4

times 4

132 = 8a + 25a

Step-by-step explanation:

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