GIVEN:

a10 = 10

S10 = -35

SOLUTION:

L = a1 + (n-1)(d)

where:

L = last term or nth term

a1 = first term

n = number of terms

d = common difference

---------

10 = a1 + (10-1)(d)

10 = a1 + 9d

a1 = 10 - 9d -------> eq.1

S = (n/2)[2(a1) + (n-1)(d)]

where:

S = sum of all terms

n = number of terms

a1 = first term

d = common difference

---------

-35 = (10/2)[2(a1) + (10-1)(d)]

-35 = (5)[2(a1) + 9d]

-35 = 10(a1) + 45d -------> eq. 2

Substitute eq. 1 to eq. 2:

-35 = 10(10 - 9d) + 45d

-35 = 100 - 90d + 45d

-35 = 100 - 45d

45d = 100 + 35

45d = 135

divide both sides by 45:

d = 3

therefore, the common difference is 3.

Substitute the value of d to eq. 1:

a1 = 10 - 9d

a1 = 10 - 9(3)

a1 = 10 - 27

a1 = -17

the value of first term is -17.