Question

Show that $6,5\dfrac{1}{2},5,4\dfrac{1}{2},4,.......$ is an A.P. Write down its:
(i) first term (ii) common difference (iii) ninth term

Answer 1

We will first find the difference between the given consecutive terms and show that it remains the same and that will be the answer of (ii) as well. Now, the first term is clearly visible in the given series and we will use the formula of ${n^{th}}$ term of the A.P. to find the ninth term.

Complete step-by-step answer:
Let us first look at the given sequence which is $6,5\dfrac{1}{2},5,4\dfrac{1}{2},4,.......$.
Let us find the difference between consecutive terms. First 6 and $5\dfrac{1}{2}$.
We know that $5\dfrac{1}{2} = \dfrac{{2 \times 5 + 1}}{2} = \dfrac{{11}}{2}$.
Hence, the difference will be $\dfrac{{11}}{2} - 6 = \dfrac{{11 - 12}}{2} = - \dfrac{1}{2}$.
Now, let us see if this is the same for further terms as well.
Consider $\dfrac{{11}}{2} - \dfrac{1}{2} = 5$ which is the next term and going on similarly, we see that it remains the same.
The first term will be clearly the first term given in the sequence. So, the first term is 6.
Since, we saw the difference to be equal to $- \dfrac{1}{2}$ for all the terms. Hence, the common difference is $- \dfrac{1}{2}$.
We know that ${a_n} = a + (n - 1)d$ , where ${a_n}$ is the ${n^{th}}$ term, a is the first term and d is the common difference.
Now putting in the given values, we will get:-
$\Rightarrow {a_9} = 6 + (9 - 1)\left( { - \dfrac{1}{2}} \right)$
On simplifying it, we will get:-
$\Rightarrow {a_9} = 6 + 8\left( { - \dfrac{1}{2}} \right) = 6 - 4 = 2$.

Hence, the ninth term is 2.

Note: The students must know that if the given sequence is not an A.P. then there can be no such thing as common term or ninth term, because we would not have any pattern related to A.P. and we cannot use their formulas as well.
The A.P. is a sequence of numbers which have the common difference.
In an A.P. if you consider any three consecutive terms, twice the second term is always equal to the sum of first and the third term.
If we increase or decrease by addition, subtraction, multiplication or division each term of A.P, the resulting sequence will also be an A.P.