Question

Which term of the A.P. 21, \[\dfrac{{62}}{3},\dfrac{{61}}{3}\]….is the first negative term?
A.64
B.63
C.62
D.65

Answer 1

Hint: Given is the problem based on the arithmetic progression. To find the negative term we will use the formula to find the nth term of that A.P. We are given the first term. We will find the common difference d from the given A.P. then we will use the formula. But only the thing is we will equate this as less than zero because it is a negative term.
Formula used:
To find nth term of the given A.P. is \[{t_n} = a + \left( {n - 1} \right)d\]
Where,
 a →is the first term
d→ is the common difference
n→ is the nth term

Complete step by step solution:
Given is an A.P.
Having first term a=21
Common difference d is given by,
  \[d = \dfrac{{61}}{3} - \dfrac{{62}}{3}\]
Since denominator is same we can directly subtract the numerators,
\[d = \dfrac{{ - 1}}{3}\]
Now the formula for the nth is,
\[{t_n} = a + \left( {n - 1} \right)d\]
But since the term is negative that is less than zero,
\[ = a + \left( {n - 1} \right)d < 0\]
Now putting the values,
\[ = 21 + \left( {n - 1} \right)\left( {\dfrac{{ - 1}}{3}} \right) < 0\]
Multiplying the brackets,
\[ = 21 - \dfrac{n}{3} + \dfrac{1}{3} < 0\]
Adding the constants using the LCM,
\[ = \dfrac{{21 \times 3 + 1}}{3} < \dfrac{n}{3}\]
Cancelling the denominator,
\[ = 63 + 1 < n\]
On adding we get,
\[ = 64 < n\]
Since n is greater than 64 only option D is correct.
 That is, the first negative term will be the 65th term.
So, the correct answer is “Option D”.

Note: Note that we can also find the common difference from the first two terms but that will take LCM to find. So we chose the second and third term because the common difference is the same between two terms of an A.P. also note that 64 is not the answer and not the other options also because since the term is negative we substituted it as less than zero and from option only term greater than 64 is 65. So that is the only option.