Question

If the third and the eighth terms of an AP are 9 and -6 respectively, which term of this AP is zero?

• A. $6^{th}$ term
• B. $5^{th}$ term
• C. $2^{nd}$ term
• D. $4^{th}$ term

Answer 1

The correct option is A $6^{th}$ term

It is given that $3^{rd}$ and $8^{th}$ term of AP are 9 and -6 respectively.

$\Rightarrow a_3=4\text{and}{a}_8=-6$.
Where, $a_3\text{and}{a}_8$ are third and eighth terms respectively.

Using formula $a_n=a+(n-1)d$ to find $n^{th}$ term of arithmetic progression, we get

$9=a+(3-1)d$
$\Rightarrow 9=a+2d$ ..................(1)

and, $-6=a+(8-1)d$
$\Rightarrow -6=a+7d$..................(2)

These are equations in two variables. Let’s solve them using the method of substitution.
Using equation $9=a+2d$ we can say that $a=9-2d$

Putting value of '$a$' in (2), we get
$-6=9-2d+7d$
$\Rightarrow -15=5d$
$\Rightarrow d=-\frac{15}{5}=-3$

Substituting '$d$' in (2), we get
$-6=a+7(-3)$
$\Rightarrow$$-6=a-21$
$\Rightarrow$$a=15$

Therefore, first term $a=15$ and Common Difference $d=-6$

We want to know which term is equal to zero.
Using formula $a_n=a+(n-1)d$ to find $n^{th}$ term of arithmetic progression, we get
$0=15+(n-1)(-3)$
$\Rightarrow 0=15-3n+3$
$\Rightarrow 0=18-3n$
$\Rightarrow 3n=18$
$\Rightarrow n=\frac{18}{3}=6$

Therefore, $6^{th}$ term of the AP is equal to 0.