Question

The 9th term of an ap is 449 and 449th term is 9 find the term which is equal to zero

Answer 1

458 is the term which is equal to zero.

Given:

$9^{t h} term = 449$

$449^{t h}  term is 9$

To find:

The term which is equal to zero = ?

Solution:

The ninth term of the AP is 449, which means that

$T_{n}=a+(n-1) d$

Putting the value of n = 9 we get, the following

$\begin{array}{l}{449=a+(9-1) d} \\ {449=a+(8) d}\end{array}$

Now the 449th term of the same AP is of value 9,

Therefore, the value after putting n = 449 we get

$9=a+(449) d$

Solving both the equation we get

$9=a+(449)d and 449=a+(8)d.$ Therefore, the value of a = 457 and d = -1

Now the term value is zero, to find which term produces the result zero, we use

$457=(n-1) \cdot 1$

Therefore, the value of n that produces term zero is 458.  

Answer 2

Answer:

$458^{th}\: term \: in \: given \\A.P \: is \: zero$

Step-by-step explanation:

Let a ,d are first term and common difference of an A.P.

_____________________

$we\: know \: that \\n^{th}\: term = a_{n}= a+(n-1)d$

_____________________

$Given ,\\9^{th}\: term = 449$

$\implies a+8d = 449\:--(1)$

$449^{th}\: term = 9$

$\implies a+448d = 9 \:---(2)$

/* Subtract equation (1) from equation (2) ,we get

$\implies 440d= -440$

$\implies d = \frac{-440}{440}$

$\implies d = -1$

/* Subtract d = -1 in equation (1) ,we get

$a+8\times (-1)=449$

$\implies a-8=449$

$\implies a = 449 +8$

$\implies a = 457$

$Let \: n^{th} \: term = 0$

$a+(n-1)d = 0$

$\implies 457+(n-1)\times (-1)=0$

$\implies 457-n+1=0$

$\implies 458-n=0$

$\implies n = 458$

Therefore,

$458^{th}\: term \: in \: given \\A.P \: is \: zero$

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