Let a_1, a_2, a_3, be in an A.P.

Question

Let $a_1, a_2, a_3, \ldots$ be in an A.P. such that $\sum_\limits{k=1}^{12} a_{2 k-1}=-\frac{72}{5} a_1, a_1 
eq 0$. If $\sum_\limits{k=1}^n a_k=0$, then $n$ is :

Accepted Answer

Given:

$    \sum\limits_{k=1}^{12} a_{2k-1} = -\frac{72}{5} a_1    $

This means:

$    a_1 + a_3 + \cdots + a_{23} = -\frac{72}{5} a_1    $

Express each term in A.P.:

The general term of the A.P. is given by $ a_k = a + (k-1)d $. Thus, for the odd indices:

$    a_1 + (a + 2d) + (a + 4d) + \cdots + (a + 22d)    $

Simplify the equation:

$    12a + 2d(1 + 2 + \cdots + 11) = -\frac{72}{5} a    $

Use the formula for sum of an arithmetic series (first 11 terms):

$    1 + 2 + \cdots + 11 = \frac{11 	imes 12}{2} = 66    $

So, the equation becomes:

$    12a + 2d(66) = -\frac{72}{5} a    $

$    12a + 132d = -\frac{72}{5} a    $

Solve for the relationship between $a$ and $d$:

$    132d = -\frac{72}{5} a - 12a     $

$    132d = -\frac{132}{5} a    $

So,

$    a = -5d \quad 	ext{(i)}    $

Second condition:

$    \sum_{k=1}^{n} a_k = 0 \quad \Rightarrow \quad S_n = 0    $

$    \frac{n}{2} [2a + (n-1)d] = 0    $

$    2a = -(n-1)d \quad 	ext{(ii)}    $

Combine equation (i) and (ii):

Substitute $ a = -5d $ into equation (ii):

$    2(-5d) = -(n-1)d     $

$    -10d = -(n-1)d    $

Therefore:

$    n-1 = 10    $

$    n = 11    $

Thus, $ n = 11 $.

Solutions · 1