IN AN AP. GIVEN d=5,s9=75 find …

Here, d = 5
S₉ = 75
We know that
{tex}{S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]{/tex}
{tex} \Rightarrow {S_9} = \frac{9}{2}\left[ {2a + (9 - 1)d} \right]{/tex}
{tex} \Rightarrow {S_9} = \frac{9}{2}\left[ {2a + 8d} \right]{/tex}
{tex} \Rightarrow {S_9} = 9\left[ {a + 4d} \right]{/tex}
{tex} \Rightarrow {S_9} = 9\left[ {a + 4 \times 5} \right]{/tex}
{tex} \Rightarrow {/tex} S₉ = 9[a + 20]
{tex} \Rightarrow {/tex} 75 = 9a + 180
{tex} \Rightarrow {/tex} 9a = 75 - 180
{tex} \Rightarrow {/tex} 9a = -105
{tex} \Rightarrow a = - \frac{{105}}{9}{/tex}
{tex} \Rightarrow a = - \frac{{35}}{3}{/tex}
Again, we know that
a_n = a + (n - 1)d
{tex} \Rightarrow {/tex} a₉ = a + (9 - 1)d
{tex} \Rightarrow {/tex} a₉ = a + 8d
{tex} \Rightarrow {a_9} = - \frac{{35}}{3} + 8(5){/tex}
{tex} \Rightarrow {a_9} = - \frac{{35}}{3} + 40{/tex}
{tex} \Rightarrow {a_9} = \frac{{ - 35 + 120}}{3}{/tex}
{tex} \Rightarrow {a_9} = \frac{{85}}{3}{/tex}