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# The 9th term of an AP is 499 and 499th term is 9. The term which is equal to zero is.A.501thB.502thC.508thD.None of these

Last updated date: 20th Jun 2024
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Hint: As given terms are in AP. We will use the formula of the nth term of an A.P i.e a+(n-1)d , where a is the first term and d is the common difference .

Let the first term of AP = a.
and the common difference = d.
given that $a_9$=499
Here n value is 9. Put value in a+(n-1)d
$a_9$ = a + 8d = 499
Therefore, a + 8d = 499 (1)
${a}_{499}$=9
${a}_{499}$ = a + 498d = 9.
Therefore, a + 498d = 9 (2)
Subtracting eq(1) from eq(2)
a+498d-a-8d=9-499
$\begin{array}{l} 490d = - 490\\ d = \dfrac{{ - 490}}{{490}} = - 1 \end{array}$
Therefore, common difference, d = -1.
Substituting the value of d in eq(1).
$\begin{array}{l} \Rightarrow a + 8d = 499\\ \Rightarrow a + \left( {8*(- 1)} \right) = 499.\\ \Rightarrow a = 499 + 8\\ \Rightarrow a = 507 \end{array}$
Therefore, first term, a = 507.
The required term = an
and an = 0
$a + (n - 1)d = 0.$
$\Rightarrow$ Putting value of a and d
$\Rightarrow 507 + (n -1 ) - 1 = 0$
$\Rightarrow 507 = n - 1$
$\Rightarrow n = 507 + 1$
$\Rightarrow n = 508$
Hence, the 508th term is equal to zero.

Note: In this type of question, use the formula to get the first and common difference terms. Then proceed with the correct formula to find the required answer.