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Q : 11        Which term of the AP : \small 3,15,27,39,...  will be \small 132 more than its  \small 54th  term?
 

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Given  AP is
\small 3,15,27,39,...
Here, a= 3
And
d= a_2-a_1 = 15 - 3 = 12
Now, let's suppose nth term of given AP is  \small 132 more than its  \small 54th  term
Then,
a_n= a_{54}+132
\Rightarrow a+(n-1)d = a+53d+132
\Rightarrow 3+(n-1)12 = 3+53\times 12+132
\Rightarrow 12n = 3+636+132+12
\Rightarrow 12n = 636+132+12
\Rightarrow n = \frac{780}{12}= 65
Therefore, 65th term of given AP is  \small 132 more than its  \small 54th  term

Posted by

Gautam harsolia

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