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Q : 1     Which term of the AP :  \small 121,117,113,..., is its first negative term?  [Hint : Find n for a_n<0 ] 

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Given AP is 
\small 121,117,113,...,
Here a = 121 \ and \ d = -4
Let suppose nth term of the AP is first negative term 
Then,
a_n = a+ (n-1)d
If nth term is negative then a_n < 0
\Rightarrow 121+(n-1)(-4) < 0
\Rightarrow 125<4n
\Rightarrow n > \frac{125}{4}=31.25
Therefore, first negative term must be 32nd term

Posted by

Gautam harsolia

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