The sum of first 'q' terms …

Let us suppose that 'a' be the first term and 'd' be the common difference of the A.P.
Now according to question it is given that

S₉ = 162.

Using formula for sum to n terms of A.P, we have
{tex}\Rightarrow \frac{9}{2}\left[ {2a + (9 - 1)d} \right] = 162{/tex}
{tex} \Rightarrow \frac{9}{2}\left[ {2a + 8d} \right] = 162{/tex}
{tex}\Rightarrow{/tex} 9a + 36d = 162 ........................(i)
Let a₆ and a₁₃ be the 6^_th and 13^th term of the A.P. respectively.
Therefore, using general form of nth term, we have,a₆ = a + 5d and a₁₃ = a + 12d
Since, {tex}\frac{{{a_6}}}{{{a_{13}}}} = \frac{1}{2}{/tex}
{tex} \Rightarrow \frac{{a + 5d}}{{a + 12d}} = \frac{1}{2}{/tex}
{tex}\Rightarrow{/tex} 2a + 10d = a + 12d
{tex}\Rightarrow{/tex} a = 2d
Now,substituting the value of a = 2d in equation (i), we get,
9(2d) + 36d = 162
{tex}\Rightarrow{/tex} 18d + 36d = 162
{tex}\Rightarrow{/tex} 54d = 162
{tex}\Rightarrow{/tex} d = 3
{tex}\Rightarrow{/tex} a = 2d = 2 {tex}\times{/tex} 3 = 6 = First term

Therefore, 15th term=a+14d=6+14(3)=48