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 $n = 10$.