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Question: 1) Is 210 a term of the A.P. 2, 6, 10, ……… If yes, then which term? 2) If \(\dfrac{4}{5},a,2\) are...

  1. Is 210 a term of the A.P. 2, 6, 10, ……… If yes, then which term?
  2. If 45,a,2\dfrac{4}{5},a,2 are three consecutive terms of A.P., then find the values of a.
  3. Which term of the A.P. 19,1815,172519,18\dfrac{1}{5},17\dfrac{2}{5} is the first negative term?
Explanation

Solution

1.) We will treat the given value as a term and then apply the formula for nth{n^{th}} term on this and see, if we get a valid value for n, this will be a term in this A.P. and will be nth{n^{th}} term only.
2.) We will find the sum of the first two terms and the last two terms and just equate them. We will thus reach the value required.
3.) We will just assume that the nth{n^{th}} term of A.P. is less than 0 and equate its formula to less than 0, then we will on putting in the values get some condition on n, which will help us to get to the required answer.

Complete step-by-step answer:
Let us first write the formula of nth{n^{th}} term which we will require in some parts of solution:
an=a+(n−1)d{a_n} = a + (n - 1)d ………..(1), where a is the first term and d is the common difference.
Part 1)
We have the AP: 2, 6, 10, ………
Here, we clearly see that the first term is given by 2. So, a = 2.
We can find the common difference by subtracting any two consecutive terms. So, here d = 6 – 2 = 4.
Now, assuming that 210 is a term of the given A.P. and applying the formula (1) on it, we will get:-
210=2+4(n−1)210 = 2 + 4(n - 1)
On simplifying it, we will get:-
⇒210=4n−2\Rightarrow 210 = 4n - 2
Rearranging it to get:-
⇒4n=212\Rightarrow 4n = 212
On solving it, we will get:-
⇒n=53\Rightarrow n = 53
We are getting a valid value of n for the term 210.
Hence, it is the fifty third term of the given A.P.
Part 2)
We know that the common difference is given by difference of two consecutive terms and it remains the same throughout the A.P., whichever consecutive terms we choose.
So, here we will get:-
⇒a−45=2−a\Rightarrow a - \dfrac{4}{5} = 2 - a
Rearranging the terms to get:-
⇒2a=2+45\Rightarrow 2a = 2 + \dfrac{4}{5}
On simplifying, we will get:-
⇒2a=145\Rightarrow 2a = \dfrac{{14}}{5}
Taking the 2 from LHS to RHS, we will get:-
⇒a=75\Rightarrow a = \dfrac{7}{5}.
Hence, the answer is a=75a = \dfrac{7}{5}.
Part 3)
Let nth{n^{th}} term be its first negative number.
So, an<0{a_n} < 0.
Here, we have the AP as: 19,1815,172519,18\dfrac{1}{5},17\dfrac{2}{5}
Hence, a = 19 and d=1815−19=915−19=−45d = 18\dfrac{1}{5} - 19 = \dfrac{{91}}{5} - 19 = - \dfrac{4}{5}
Now, using (1), we will get:-
an=a+(n−1)d<0{a_n} = a + (n - 1)d < 0
Putting in the values will give us:-
⇒19−(n−1)45<0\Rightarrow 19 - (n - 1)\dfrac{4}{5} < 0
On simplifying it, we will get:-
⇒19−45n+45<0\Rightarrow 19 - \dfrac{4}{5}n + \dfrac{4}{5} < 0
On simplifying it, we will get:-
⇒−45n<\-995\Rightarrow - \dfrac{4}{5}n < \- \dfrac{{99}}{5}
On simplifying it further, we will get:-
⇒n>994=24.75\Rightarrow n > \dfrac{{99}}{4} = 24.75.
Hence, the twenty fifth term of the A.P. will be the first negative number.

Note: The students must note that in the second part, we basically used the difference of the first two and last two terms to give the same result that is happening in AP because if the first term is a, then a + d and then a+ 2d. We can note that a+ d – a = d = a+ 2d – a – d.
And in the third part, always remember that if we multiply an inequality by a negative sign, the sign changes.