There is n arithmetic means between 2 and 46 such that the first mean: last mean is 1:7. find the value of n.

Step-by-step explanation:

The nth term of an arithmetic sequence is expressed as:

Tn  = a + (n-1)d

a is the first term

n is the number of terms

d is the common difference

Let the sequence be 2, T1, T2, T3...Tn, 46

The common difference is expressed as d = b-a/n+1

T1 = 2 + d

Given b = 46 and a = 2

d = 46-2/n+1

d = 44/n+1

T1 = 2 + (44/n+1)

T1 = 2(n+1)+44/n+1

T1 = 2n+2+44/n+1

T1 = 2n+46/n+1

Similarly;

Tn =  2 + nd

Tn = 2 + n(44/n+1)

Tn = 2 + 44n/n+1

Tn = 2(n+1)+44n/n+1

Tn = (2n+2+44n)/n+1

Tn = 46n+2/n+1

If the ratio of the first mean to last mean is 1:7 then T1/Tn = 1/7

(2n+46/n+1)÷(46n+2/n+1) = 1/7

2n+46/n+1 * n+1/46n+2 = 1/7

2n+46/46n+2 = 1/7

Cross multiply and solve for n:

46n + 2 = 7(2n + 46)

46n + 2 = 14n + 322

46n - 14n = 322-2

32n = 320

n = 320/32

n = 10

Hence the value of n is 10

What are the answers again??

Step-by-step explanation: