Answer 1

Given that, a₃ = 15, S₁₀ = 125

As a_n = a + (n − 1)d,

a₃ = a + (3 − 1)d

15 = a + 2d ... (i)

S_n = `n/2 [2a + (n - 1)d]`

S₁₀ = `10/2 [2a + (10 - 1)d]`

125 = 5(2a + 9d)

25 = 2a + 9d ... (ii)

On multiplying equation (i) by (ii), we get

30 = 2a + 4d ... (iii)

On subtracting equation (iii) from (ii), we get

−5 = 5d

d = −1

From equation (i),

15 = a + 2(−1)

15 = a − 2

a = 17

a₁₀ = a + (10 − 1)d

a₁₀ = 17 + (9) (−1)

a₁₀ = 17 − 9 = 8