How many terms are there in the A.P. whose first and fifth terms are −14 and 2 respectively and the sum of the terms is 40?
Solution
\[\text { We have: } \]
\[ a = - 14 \text { and } S_n = 40 . . . (i)\]
\[ a_5 = 2\]
\[ \Rightarrow a + \left( 5 - 1 \right)d = 2\]
\[ \Rightarrow - 14 + 4d = 2\]
\[ \Rightarrow 4d = 16\]
\[ \Rightarrow d = 4 . . . (ii)\]
\[\text { Also }, S_n = \frac{n}{2}\left[ 2a + (n - 1)d \right]\]
\[ \Rightarrow 40 = \frac{n}{2}\left[ 2\left( - 14 \right) + (n - 1) \times 4 \right] (\text { From }(i) \text { and } (ii))\]
\[ \Rightarrow 80 = n\left[ - 28 + 4n - 4 \right]\]
\[ \Rightarrow 80 = 4 n^2 - 32n\]
\[ \Rightarrow n^2 - 8n - 20 = 0\]
\[ \Rightarrow (n - 10)(n + 2) = 0\]
\[ \Rightarrow n = 10, - 2\]
\[\text { But, n cannot be negative } . \]
\[ \therefore n = 10 \]
