Question

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

Answer 1

Hint: The nth term of an arithmetic progression is given as ${{T}_{n}}=a+\left( n-1 \right)d$ where ‘a’ and ‘d’ are the first time and common difference of the A.P. and ${{T}_{n}}$ is denoting the nth term. Form the equations using the given conditions and solve it further to get the answer.

Complete step-by-step answer:
As we know that general term of an A.P. is given as ${{T}_{n}}=a+\left( n-1 \right)d$where a, d are denoting the first term of the A.P. and common difference respectively and ${{T}_{n}}$is denoting the nth term.
So, now coming to the question, we have the 3rd and 9th terms of the A.P. as 4 and -8 respectively.
Hence, we can form two equations with the help of the relation of the nth term of any A.P.
Let common difference and first term be d and a respectively of A.P. , hence
${{T}_{n}}=a+\left( n-1 \right)d$………….. (i)
So, 3rd term can be written as
${{T}_{3}}=a+\left( 3-1 \right)d=a+2d$
As we know ${{T}_{3}}=4$, hence, we get
${{T}_{3}}=a+2d=4$……….. (ii)
Similarly, 9th term can be written with the help of equation (i) as
${{T}_{g}}=a+\left( 9-1 \right)d=a+8d$
As we know ${{T}_{9}}=-8$, hence, we get
a + 8d = -8 ………... (iii)
Now, subtract equation (ii) and (iii) to get values of a and d. hence, we get
(a + 2d) – (a + 8d) = 4 – (-8)
a + 2d – a – 8d = 4 + 8
a – a + 2d – 8d = 12
-6d = 12
d = -2 ……………. (iv)
Now, put the value of ‘d’ from equation (iv) in equation (iii) to get the value of ‘a’. Hence, we get
a + 8 (-2) = -8
a – 16 = -8
a = 16 – 8 = 8
a = 8 ………….. (v)
Hence, the given A.P. in the question will have first term as 8 and common difference as ‘-2’.
Now, we need to determine which term will be zero of the A.P.
Let ${{t}^{th}}$ term of the given A.P. will be O.
Hence, we can write the ${{t}^{th}}$ term from the equation (i) as
${{T}_{t}}=a+\left( t-1 \right)d$
Where a = 8 from equation (v), d = -2 from equation (iv) and ${{T}_{t}}=0$
So, we get
0 = 8 + (t – 1) (-2)
-8 = (t -1) (-2)
 (t -1) = 4
t = 5
Hence, the 5th term of the A.P. will be 0.

Note: One can solve the equations a + 2d = 4 and a + 8d = -8 by substitution as well. We can take the value of ‘a’ from one equation and put it in the second equation.
One can go wrong with the nth term of the A.P. as well. One may write the 3rd and 9th terms as (a + 3d) and (a + 9d) in case of very fast calculations, which are wrong. So, always go by basics and be clear with the terms of ${{T}_{n}}=a+\left( n-1 \right)d$ to get the correct equations.