From the question,
First term a = \[34\],
Difference d = \[32\text{ }\text{ }34\text{ }=\text{ }-2\]
So, common difference d = \[-2\]
Last term Tn = 10
We know that, \[{{T}_{n}}\] = a + (n – 1)d
\[\begin{array}{*{35}{l}}
10\text{ }=\text{ }34\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)\left( -2 \right) \\
-24\text{ }=\text{ }\text{ }2\left( n\text{ }\text{ }1 \right) \\
-24\text{ }=\text{ }\text{ }2n\text{ }+\text{ }2 \\
2n\text{ }=\text{ }24\text{ }+\text{ }2 \\
2n\text{ }=\text{ }26 \\
n\text{ }=\text{ }26/2 \\
n\text{ }=\text{ }13 \\
\end{array}\]
\[\begin{array}{*{35}{l}}
{{S}_{n}}~=\text{ }n/2\left( a\text{ }+\text{ }1 \right) \\
=\text{ }13/2\text{ }\left( 34\text{ }+\text{ }10 \right) \\
=\text{ }13/2\text{ }\left( 44 \right) \\
=\text{ }13\text{ }\left( 22 \right) \\
=\text{ }286 \\
\end{array}\]
(ii) \[\text{ }\mathbf{5}\text{ }+\text{ }\left( \text{ }\text{ }\mathbf{8} \right)\text{ }+\text{ }\left( \text{ }\text{ }\mathbf{11} \right)\text{ }+\text{ }\ldots \text{ }+\text{ }\left( \text{ }\text{ }\mathbf{230} \right)\]
Solution:-
From the question,
First term a = \[-5\],
Difference d = \[-8\text{ }\text{ }\left( -5 \right)\text{ }=\text{ }-8\text{ }+\text{ }5\text{ }=\text{ }-3\]
So, common difference d = \[-3\]
Last term \[{{T}_{n}}~=\text{ }-230\]
We know that, \[{{T}_{n}}\] = a + (n – 1)d
\[\begin{array}{*{35}{l}}
-230\text{ }=\text{ }-5\text{ }+\text{ }\left( n\text{ }\text{ }1 \right)\left( -3 \right) \\
-230\text{ }=\text{ }\text{ }5\text{ }\text{ }3n\text{ }+\text{ }3 \\
-230\text{ }=\text{ }\text{ }2\text{ }\text{ }3n \\
3n\text{ }=\text{ }230\text{ }-2 \\
3n\text{ }=\text{ }228 \\
n\text{ }=\text{ }228/3 \\
n\text{ }=\text{ }76 \\
\end{array}\]
Therefore,
\[\begin{array}{*{35}{l}}
{{S}_{n}}~=\text{ }n/2\text{ }\left( a\text{ }+\text{ }l \right) \\
=\text{ }76/2\text{ }\left( -5\text{ }+\text{ }\left( -230 \right) \right) \\
=\text{ }38\text{ }\left( -5\text{ }\text{ }230 \right) \\
=\text{ }38\text{ }\left( 235 \right) \\
=\text{ }\text{ }8930 \\
\end{array}\]