Find the first term and common difference of an AP, whose 7th term is 29/6 and 15th term is 11/2.

The first term is 13/3 and the common difference is d = 1/12

The formula is a(n) = 13/3 + (1/12)(n - 1)        

Step-by-step explanation:

The general equation for an arithmetic progression is:

a(n) = a(1) + d(n - 1), where d is the common difference/

Case 1:  n = 7:  29/6 = a(1) + d(7 - 1), or 29/6 = a(1) + d(6)

Case 2:  n = 15:  11/2 = a(1) + d(15 - 1) =  a(1) + d(14)

Then our system of linear equations is:

a(1) + 6d = 29/6

a(1) + 14d = 11/2

Let's solve this by elimination by addition and subtraction.  Subtract the first equation from the second.  We get:

Substituting 1/12 for d in the first equation, we get:

a(1) + 14(1/12) = 11/2 or 66/12 (using the LCD 12)

Then a(1) = 66/12 - 14/12 = 52/12 = 13/3

The first term is 13/3 and the common difference is d = 1/12

The arithmetic sequence formula for this problem is thus:

a(n) = 13/3 + (1/12)(n - 1)                                                                          8

step-by-step explanation:

because 1/3 is a fraction. we can tell because there is a divisor symbol between the 1 and the 3. if we were to add 2 fractions together, we would add the numerators, which is the number on top, but not the denominators, which is the number on the bottom

1/3+1/3=2/3

hope this !