Question

If the 5th and 12th terms of an A.P. are 30 and 65 respectively, what is the sum of first 20 terms?

Answer 1

\[\text { We have }: \]

\[ a_5 = 30\]

\[ \Rightarrow a + \left( 5 - 1 \right)d = 30\]

\[ \Rightarrow a + 4d = 30 . . . (i)\]

\[\text { Also }, a_{12} = 65\]

\[ \Rightarrow a + \left( 12 - 1 \right)d = 65\]

\[ \Rightarrow a + 11d = 65 . . . . . (ii)\]

\[\text { Solving (i) and (ii), we get }: \]

\[7d = 35\]

\[ \Rightarrow d = 5\]

\[\text { Putting the value of d in (i), we get }: \]

\[a + 4 \times 5 = 30\]

\[ \Rightarrow a = 10\]

\[ \therefore S_{20} = \frac{20}{2}\left[ 2 \times 10 + (20 - 1) \times 5 \right]\]

\[ \Rightarrow S_{20} = 10\left[ 2 \times 10 + (20 - 1) \times 5 \right]\]

\[ \Rightarrow S_{20} = 1150\]