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Question
If the 5th and 12th terms of an A.P. are 30 and 65 respectively, what is the sum of first 20 terms?
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Solution
\[\text { We have }: \]
\[ a_5 = 30\]
\[ \Rightarrow a + \left( 5 - 1 \right)d = 30\]
\[ \Rightarrow a + 4d = 30 . . . (i)\]
\[\text { Also }, a_{12} = 65\]
\[ \Rightarrow a + \left( 12 - 1 \right)d = 65\]
\[ \Rightarrow a + 11d = 65 . . . . . (ii)\]
\[\text { Solving (i) and (ii), we get }: \]
\[7d = 35\]
\[ \Rightarrow d = 5\]
\[\text { Putting the value of d in (i), we get }: \]
\[a + 4 \times 5 = 30\]
\[ \Rightarrow a = 10\]
\[ \therefore S_{20} = \frac{20}{2}\left[ 2 \times 10 + (20 - 1) \times 5 \right]\]
\[ \Rightarrow S_{20} = 10\left[ 2 \times 10 + (20 - 1) \times 5 \right]\]
\[ \Rightarrow S_{20} = 1150\]
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