The forty and the 9th term of an AP is -3 and 12 respectively. Find the common difference and some of the frist 7th term

term(40) = a+39d = -3 **

term(9) = a+8d = 12 ***
subtract them:
31d = -15
d = -15/31
back in ***
a + 8(-15/31) = 12
a = 492/31

"and some of the frist 7th term"
I will assume that is supposed to mean:
find the sum of the first 7 terms

sum(7) = (7/2)( 984/31 + 6(-15/31) ) = 3129/31

I was expecting more civilized numbers.
Are there any typos other than your grammar and spelling ones ?

If the initial term of an arithmetic progression is a a1 and the common difference of successive members is d, then the nth term of the AP is given by:

an = a1 + ( n - 1 ) d

a4 = a1 + ( 4 - 1 ) d

a4 = a1 + 3 d = - 3

a9 = a1 + ( 9 - 1 ) d

a9 = a1 + 8 d = 12

Now you must solve system of 2 equations with 2 unknow:

a1 + 3 d = - 3 and a1 + 8 d = 12

a1 + 3 d = - 3
-
a1 + 8 d = 12
_______________

a1 - a1 + 3 d - 8 d = - 3 - 12

0 - 5 d = - 15

- 5 d = - 15 divide both sides by - 5

d = - 15 / - 5 = 3

Replace this value in equation:

a1 + 3 d = - 3

a1 + 3 * 3 = - 3

a1 + 9 = - 3 Subtract 9 to both sides

a1 + 9 - 9 = - 3 - 9

a1 = - 12

an = a1 + ( n - 1 ) d

a7 = a1 + ( 7 - 1 ) d

a7 = a1 + 6 d

a7 = - 12 + 6 * 3

a7 = - 12 + 18

a7 = 6

Your AP:

-12 , - 9 , - 6 , - 3 , 0 , 3 , 6 , 9 , 12 , 15 , 18...

To find the common difference of an arithmetic progression (AP), we can use the formula:

nth term = a + (n - 1)d

where:
nth term is the term you want to find,
a is the first term,
n is the position of the term,
d is the common difference.

We are given that the 40th term is -3 and the 9th term is 12.

Plugging these values into the formula, we can set up two equations:

-3 = a + (40 - 1)d -------- (1)
12 = a + (9 - 1)d -------- (2)

Simplifying equation (1):

-3 = a + 39d

Simplifying equation (2):

12 = a + 8d

We now have a system of linear equations. We can solve this system to find the values of a and d.

Subtracting equation (2) from equation (1):

-3 - 12 = a + 39d - a - 8d
-15 = 31d

Dividing both sides by 31:

d = -15/31

Therefore, the common difference is -15/31.

To find the sum of the first seven terms, we can use the formula:

Sum of n terms = (n/2)(2a + (n - 1)d)

Substituting n = 7, a = first term, and d = -15/31:

Sum of 7 terms = (7/2)(2a + (7 - 1)(-15/31))
= (7/2)(2a - 6(15/31))
= (7/2)(2a - 90/31)
= (7/2)(2a - 45/31)
= (7/2)((62a - 45)/31)
= (7/2)(62a - 45)/31

Since we don't have the value of the first term (a), we cannot calculate the exact sum of the first seven terms.

To find the common difference in an arithmetic progression (AP), we can use the formula:

an = a + (n-1)d
where:
an = the nth term of the AP
a = the first term of the AP
n = the term number
d = the common difference

Given the 40th term (a40) is -3 and the 9th term (a9) is 12, we can use these values to solve for the common difference (d).

Using the formula:
a40 = a + (40-1)d
-3 = a + 39d --------(1)

a9 = a + (9-1)d
12 = a + 8d --------(2)

Now, we have a system of two equations with two unknowns (a and d). We can solve these equations simultaneously to find the values of a and d.

From equation (2), we can express a in terms of d:
a = 12 - 8d

Substituting this expression for a into equation (1), we get:
-3 = (12 - 8d) + 39d

Simplifying the equation:
-3 = 12 + 31d

Rearranging the terms:
31d + 12 = -3

Subtracting 12 from both sides:
31d = -15

Dividing both sides by 31:
d = -15/31

So, the common difference (d) between the terms in the arithmetic progression is -15/31.

To find the sum of the first 7 terms of the AP, we can use the formula for the sum of an AP:
Sn = (n/2)(2a + (n-1)d)
where:
Sn = sum of the first n terms of the AP

Substituting n = 7, a = -3, and d = -15/31 into the formula, we get:
S7 = (7/2)(2(-3) + (7-1)(-15/31))
= (7/2)(-6 - (6)(15/31))

Evaluating the expression:
S7 = (7/2)(-6 - 90/31)
= (7/2)(-186/31)
= -651/31

Therefore, the sum of the first 7 terms of the AP is -651/31.