Question

The sum of three consecutive terms of an AP is 21 and the sum of the squares of these terms is 165. Find these terms

Answer 1

Let the required terms be (a - d), a and (a + d).

Then (a - d) + a + (a + d) = 21

\(\Rightarrow\) 3a = 21

\(\Rightarrow\) a = 7

Also, (a - d)² + a² + (a + d)² = 165

\(\Rightarrow\) 3a² + 2d² = 165

\(\Rightarrow\) (3 x 49 + 2d²) = 165

\(\Rightarrow\) 2d² = 165 - 147 = 18

\(\Rightarrow\) d² = 9

\(\Rightarrow\) d = \(\pm3\)

Thus, a = 7 and d = \(\pm3\)

Hence, the required terms are (4,7,10) or (10,7,4).