The sum of the third and …

Let the first term and the common difference of the AP be a and d respectively.
According to the question,
Third term + seventh term = 6
{tex} \Rightarrow {/tex} [a + (3 - 1)d] + [a + (7 - 1)d] = 6 = a + (n - 1)d
{tex} \Rightarrow {/tex} (a + 2d) + (a + 6d) = 6 {tex} \Rightarrow {/tex} 2a + 8d = 6
{tex} \Rightarrow {/tex} a + 4d = 3 ..... (1)
Dividing throughout by 2 &
(third term) (seventh term) = 8
{tex} \Rightarrow {/tex} (a + 2d) (a + 6d) = 8
{tex} \Rightarrow {/tex} (a + 4d - 2d) (a + 4d + 2d) = 8
{tex} \Rightarrow {/tex} (3 - 2d) (3 + 2d) = 8
{tex} \Rightarrow {/tex} 9 - 4d² = 8
{tex} \Rightarrow 4{d^2} = 1 \Rightarrow {d^2} = \frac{1}{4} \Rightarrow d + \pm \frac{1}{2}{/tex}
Case I, when {tex}d = \frac{1}{2}{/tex}
Then from (1), {tex}a + 4\left( {\frac{1}{2}} \right) = 3{/tex}
{tex} \Rightarrow {/tex} a + 2 = 3 {tex} \Rightarrow {/tex} a = 3 - 2 {tex} \Rightarrow {/tex} a = 1
{tex}\therefore {/tex} Sum of first sixteen terms of the AP = S₁₆
{tex} = \frac{{16}}{2}[2a + (16 - 1)d]{/tex} {tex}\because {S_n} = \frac{n}{2}[2a + (n - 1)d]{/tex}
= 8[2a + 15d]
{tex} = 8[2(1) + 15(\frac{1}{2})]{/tex}
{tex} = 8[12 + \frac{{15}}{2}]{/tex}
{tex} = 8[\frac{{19}}{2}]{/tex}
{tex} = 4 \times 19 = 76{/tex}
Case II. When {tex}d = - \frac{1}{2}{/tex}
Then from (1),
{tex}a + 4\left( { - \frac{1}{2}} \right) = 3{/tex}
{tex} \Rightarrow {/tex} a - 2 = 3 {tex} \Rightarrow {/tex} a = 3 + 2 {tex} \Rightarrow {/tex} a = 5
{tex}\therefore {/tex} Sum of first sixteen terms of the AP = S₁₆
{tex} = \frac{{16}}{2}[2a + (16 - 1)d]{/tex} {tex}\because {S_n} = \frac{n}{2}[2a + (n - 1)d]{/tex}
{tex} = 8[2a + 15d] = 8\left[ {2(5) + 15\left( { - \frac{1}{2}} \right)} \right] = 8\left[ {10 - \frac{{15}}{2}} \right] = 8\left[ {\frac{5}{2}} \right] = 20{/tex}