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 arithmetic progression is \( 27, 24, 21, \ldots \). The first term is \( a = 27 \) and the common difference is \( d = -3 \).

The formula for the sum of the first \( n \) terms of an A.P. is:

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

Given \( S_n = 105 \), we substitute the values:

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

Simplification yields:

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

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

Multiplying by 2:

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

This leads to the quadratic equation:

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

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

Dividing by 3:

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

Solving for \( n \), we find:

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

Thus, \( n = 7 \) results in a sum of 105.

To determine the term that is zero, we use the formula for the \( n \)-th term:

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

Setting \( a_n = 0 \):

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

Solving this equation:

\[ 0 = 27 - 3n + 3 \]

\[ 0 = 30 - 3n \]

\[ 3n = 30 \]

\[ n = 10 \]

Therefore, the 10th term is zero.