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Question
If 12th term of an A.P. is −13 and the sum of the first four terms is 24, what is the sum of first 10 terms?
Solution
Let a be the first term and d be the common difference.
\[a_{12} = - 13\]
\[ \Rightarrow a + \left( 12 - 1 \right)d = - 13\]
\[ \Rightarrow a + 11d = - 13 . . . (i)\]
\[\text { Also, } S_4 = 24\]
\[ \Rightarrow \frac{4}{2}\left[ 2a + (4 - 1)d \right] = 24\]
\[ \Rightarrow 2\left( 2a + 3d \right) = 24\]
\[ \Rightarrow 2a + 3d = 12 . . . (ii) \]
\[\text { From (i) and (ii), we get }: \]
\[19d = - 38\]
\[ \Rightarrow d = - 2\]
\[\text { Putting the value of d in (i), we get }: \]
\[a + 11\left( - 2 \right) = - 13\]
\[ \Rightarrow a = 9\]
\[ S_{10} = \frac{10}{2}\left[ 2 \times 9 + (10 - 1)\left( - 2 \right) \right]\]
\[ \Rightarrow S_{10} = 5\left[ 18 + 9\left( - 2 \right) \right] = 0\]
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