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

Answer 1

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 ?

Answer 2

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...