The sum of the third and …

Let the AP be a - 4d, a - 3d, a - 2d, a - d, a, a + d, a + 2d, a + 3d,...
Then, a₃ = a - 2d, a₇ = a + 2d
{tex} \Rightarrow {/tex} a₃ + a₇ = a - 2d + a + 2d = 6
{tex} \Rightarrow {/tex} 2a = 6 {tex} \Rightarrow {/tex} a = 3 ..... (i)
Also (a - 2d) (a + 2d) = 8
{tex} \Rightarrow {/tex} a² - 4d² = 8 {tex} \Rightarrow {/tex} 4d² = a² - 8 {tex} \Rightarrow {/tex} 4d² = 3² - 8
{tex} \Rightarrow {/tex} 4d² = 1 {tex} \Rightarrow {d^2} = \frac{1}{4} \Rightarrow d = \pm \frac{1}{2}{/tex}
Taking {tex}d = \frac{1}{2},{/tex}
{tex}{S_{16}} = \frac{{16}}{2}\left[ {2 \times (a - 4d) + (16 - 1)d} \right]{/tex}{tex} = 8\left[ {2 \times \left( {3 - 4 \times \frac{1}{2}} \right) + 15 \times \frac{1}{2}} \right]{/tex}
{tex} = 8\left[ {2 + \frac{{15}}{2}} \right] = 8 \times \frac{{19}}{2} = 76{/tex}
Taking {tex}d = \frac{{ - 1}}{2}{/tex},
{tex}{S_{16}} = \frac{{16}}{2}\left[ {2 \times (a - 4d) + (16 - 1)d} \right]{/tex}{tex} = 8\left[ {2 \times \left( {3 - 4 \times \frac{1}{2}} \right) + 15 \times \frac{1}{2}} \right]{/tex}
{tex} = 8\left[ {\frac{{20 - 15}}{2}} \right] = 8 \times \frac{5}{2} = 20{/tex}
{tex}\therefore {/tex} S₁₆ = 20 and 76