Question

How many terms of the $A.P.$ $27, 24, 21, . . .$ must be taken so that their sum is 105? Which term of the A.P. is zero?

Answer 1

- The given arithmetic progression is 27, 24, 21, ..., with the first term \( a = 27 \) and the common difference \( d = -3 \).
- The sum of the first \( n \) terms of an A.P. is given by:

\[ S_n = \frac{n}{2} [2a + (n-1)d] \]

- Substituting the known values:

\[ 105 = \frac{n}{2} [2(27) + (n-1)(-3)] \]

Simplifying:

\[ 105 = \frac{n}{2} [54 - 3n + 3] \] \[ 105 = \frac{n}{2} (57 - 3n) \]

Multiplying both sides by 2:

\[ 210 = n(57 - 3n) \]

Solving the quadratic equation:

\[ 210 = 57n - 3n^2 \] \[ 3n^2 - 57n + 210 = 0 \]

Dividing by 3:

\[ n^2 - 19n + 70 = 0 \]

Solving for \( n \):

\[ n = 7 \text{ or } n = 10 \]

- Therefore, \( n = 7 \) gives the sum as 105.
- To find the term that is zero, we use the formula for the \( n \)-th term:

\[ a_n = a + (n-1)d = 27 + (n-1)(-3) = 0 \]

Solving:

\[ 27 + (n-1)(-3) = 0 \] \[ 27 - 3n + 3 = 0 \] \[ 30 = 3n \] \[ n = 10 \]

So, the term is zero at \( n = 10 \).