In the equal difference series an, a1 1, d 3 is known in the equal ratio series bn, b1 2, q 2, then

1) an=a1+(n-1)d=1+(n-1)*3=3n-2bn=b1*q^(n-1)=2*2^(n-1)=2^n2) b4=2^4=16

If it is in, then 3n-2=16 3n=18 n=9 is the 9th term.

3) b6=a22=64

b8=86=256

b10=a342=1024

bn=2＾an

bn+1=2 lead stool an+1

bn+1 bn=2 an+1 an=2 an=2 (an+1-an)=2 d

bn) is a proportional series.

The number series {an} and {bn} are both equal difference series.

Let the tolerance of the series {an} be da, and the tolerance of {bn} is db1+b1=7

a3+b3=a1+2da+b1+2db=7+2(da+db)=21da+db=7

a5+b5a1+4da+b1+4db

a1+b1+4(da+db)

The sum of the tolerances of the two series is 7, so 28+7=35

c1=1=a1+b1 b1=0 a1=1

c2=1=a2+b2 a1q+b1+d=1 q+d=1c3=2=a3+b3=a1q 2+b1+2d q 2+2d=2 q=2 d=-1

The sum of the first 10 terms of the sequence {cn} = the sum of the first 10 terms of the sequence {an} + the sum of the first 10 terms of the series {bn}.

The sum of the first ten terms of the number series {evoking the skin an} of the first ten terms of the chain = a1 (1-2 10) (1-2) = 2 10-1 = 511

Wait, I'll do the math.

a3+a5=2a4=b4

i.e. 2+6d=q 3---1).

Because b2b3=a8

That is, q*q 2=1+7d

q^3=1+7d---2)

1) (2) Synopids d=1, q=2

So s10 = (1 + 10) * 10 2 = 55

t10=(2^10-1)/(2-1)=1023

Let the tolerance be d and the tolerance be q

a2=b2a1+d=b1q

a1 = 1, b1 = 1 substitution, get.

d+1=qd=q-1

2a3-b3=1

2(a1+2d)-b1q²=1

a1=1, b1=1 substitution, arrangement, getting.

d=(q²-1)/4

q-1=(q²-1)/4

Finish, get (q-1)(q-3)=0 get q=1 or q=3(1) when q=1.

d=q-1=1-1=0

an=a1+(n-1)d=1+0·(n-1)=1bn=b1qⁿ⁻¹=1·1ⁿ⁻¹=1

The general formula for the series is an=1, and the formula for the series is bn=1(2) when q=3.

d=q-1=1-1=2

an=a1+(n-1)2=1+2· (n-1)=2n-1bn=b1q =1·3 =3 The general formula for the series is an=2n-1, and the general formula for the series is bn=3

Let the tolerance d and the tolerance q be brought to the last two formulas respectively, and you can find them.