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 $\sum _{k=1}^{12}{a}_{2k-1}=-\frac{72}{5}{a}_{1},{a}_{1}
e 0$. If $\sum _{k=1}^{n}{a}_{k}=0$, then $n$ is :

Accepted Answer

Given:

$\sum _{k=1}^{12}{a}_{2k-1}=-\frac{72}{5}{a}_{1}$

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,

$a=-5d\,$

Second condition:

$\sum _{n}^{k=1}{a}_{k}=0\, \Rightarrow \, {S}_{n}=0$

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

$2a=-(n-1)d\,$

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