Question

If the third and the ninth terms of an AP are 4 and -8 respectively, which term of this AP is zero?

• A. 5
• B. 4
• C. 3
• D. 6

Answer 1

The correct option is A

5

It is given that $3^{rd}$ and $9^{th}$ term of AP are 4 and -8 respectively.
It means $a_3=4and{a}_9=-8$.
Where, $a_{3}and{a}_9$ are third and ninth terms respectively.
Using formula $a_n=a+(n-1)d$ to find $n^{th}$ term of arithmetic progression, we get
$4=a+(3-1)d$
$\Rightarrow 4=a+2d$
$-8=a+(9-1)d$
$\Rightarrow -8=a+8d$
These are equations in two variables. Let’s solve them using the method of substitution.
Using equation $4=a+2d$ we can say that $a=4-2d$
Putting value of a in other equation i.e. $-8=a+8d$ we get
$-8=4-2d+8d$
$\Rightarrow -12=6d$
$\Rightarrow d=-\frac{12}{6}=-2$
Putting value of d in equation: $-8=a+8d$, we get
$-8=a+8(-2)$
$\Rightarrow$$-8=a-16$
$\Rightarrow$$a=8$
Therefore, first term a = 8 and Common Difference d =−2
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
$0=8+(n-1)(-2)$
$\Rightarrow 0=8-2n+2$
$\Rightarrow 0=10-2n$
$\Rightarrow 2n=10$
$\Rightarrow n=\frac{10}{2}=5$

Therefore, $5^{th}$ term is equal to 0.