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, ... with first term \( a = 27 \) and common difference \( d = -3 \). The sum of the first \( n \) terms is \( S_n = \frac{n}{2} [2a + (n-1)d] \). Given \( S_n = 105 \):

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

Simplifying the equation:

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

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

Multiply by 2:

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

Rearrange into a quadratic equation:

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

Divide by 3:

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

Solving for \( n \):

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

Thus, \( n = 7 \) yields a sum of 105. To find the term that is zero, use the formula \( a_n = a + (n-1)d \):

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

Solving for \( n \):

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

\[ 30 = 3n \]

\[ n = 10 \]

The 10th term is zero.