a3 =15,s10=125, find d and a10

Here, a₃ = 15
S₁₀ = 125
We know that
a_n = a + (n - 1)d
{tex} \Rightarrow {/tex} a₃ = a + (3 - 1)d
{tex} \Rightarrow {/tex} a₃ = a + 2d
{tex} \Rightarrow {/tex} 15 = a + 2d
{tex} \Rightarrow {/tex} a + 2d = 15 ...... (1)
Again, we know that
{tex}{S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]{/tex}
{tex} \Rightarrow {S_{10}} = \frac{{10}}{2}\left[ {2a + (10 - 1)d} \right]{/tex}
{tex} \Rightarrow {/tex} S₁₀ = 5(2a + 9d)
{tex} \Rightarrow {/tex} 125 = 5(2a + 9d)
{tex} \Rightarrow {/tex} 25 = 2a + 9d
{tex} \Rightarrow {/tex} 2a + 9d = 25 ....... (2)
Solving equation (1) and equation (2), we get
a = 17
d = -1
Now an = a + (n - 1)d
{tex} \Rightarrow {/tex} a₁₀ = a + (10 - 1)d
{tex} \Rightarrow {/tex} a₁₀ = a + 9d
{tex} \Rightarrow {/tex} a₁₀ = 17 + 9(-1)
{tex} \Rightarrow {/tex} a₁₀ = 17 - 9
{tex} \Rightarrow {/tex} a₁₀ = 8