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Question
In an A.P. the first term is 2 and the sum of the first five terms is one fourth of the next five terms. Show that 20th term is −112.
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Solution
\[\text { Given }: \]
\[ a = 2, S_5 = \frac{1}{4}\left( S_{10} - S_5 \right)\]
\[\text { We have: } \]
\[ S_5 = \frac{5}{2} \left[ 2 \times 2 + (5 - 1)d \right]\]
\[ \Rightarrow S_5 = 5\left[ 2 + 2d \right] . . . . (i)\]
\[\text { Also }, S_{10} = \frac{10}{2}\left[ 2 \times 2 + (10 - 1)d \right]\]
\[ \Rightarrow S_{10} = 5\left[ 4 + 9d \right] . . . . . (ii)\]
\[ \because S_5 = \frac{1}{4}\left( S_{10} - S_5 \right) \]
\[\text { From (i) and (ii), we have: } \]
\[ \Rightarrow 5\left[ 2 + 2d \right] = \frac{1}{4}\left[ 5(4 + 9d) - 5(2 + 2d) \right]\]
\[ \Rightarrow 8 + 8d = 4 + 9d - 2 - 2d\]
\[ \Rightarrow d = - 6\]
\[ \therefore a_{20} = a + \left( 20 - 1 \right)d\]
\[ \Rightarrow a_{20} = a + 19d\]
\[ \Rightarrow a_{20} = 2 + 19\left( - 6 \right)\]
\[ \Rightarrow a_{20} = - 112\]
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