Question

The 3rd term of an arithmetic series is 12 and the 10th term is -93.
a Find the first term and the common difference.
b Find the term in which the series turns negative.​

Answer 1

Step-by-step explanation:

________❤HOLA❤_______

GIVEN:

T3 = 12

T10= -93

LET a be the first term and d be the common difference.

we know that;

a + (n-1)d = Tn

so,

a + (3-1)d = 12

a+ 2d = 12 --------(1)

a+(10-1)d = -93

a+9d = -93 ---------(2)

now,

subtracting eq 1 and 2,

a + 2d = 12

a + 9d = -93

(-) (-) (+)

_______________

-7d = 105

d = 105/-7

d = -15

hence, d = -15

putting the value of d in eq, 1

a + 2 (-15)= 12

a -30 =12

a = 42

so, first term is 42

common difference is - 15.

let nth term will be first negative term

now,

Tn < 0

a +(n-1)d <0

42+(n-1)-15 <0

42 -15n +15 <0

57-15n <0

57 <15 n

57/15 <n