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Q : 3     In the following APs, find the missing terms in the boxes :

             (iii)   \small 5,\: \fbox { },\: \fbox { },\: 9\frac{1}{2}

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Given AP series is
\small 5,\: \fbox { },\: \fbox { },\: 9\frac{1}{2}
Here, a = 5 , n = 4 \ and \ a_4 = 9\frac{1}{2}= \frac{19}{2}
Now, we know that
a_n = a+(n-1)d
\Rightarrow a_4 =5+(4-1)d
\Rightarrow \frac{19}{2} -5=3d
\Rightarrow d = \frac{19-10}{2\times 3} = \frac{9}{6} = \frac{3}{2}
Now,
a_2= a_1+d
a_2 = 5+\frac{3}{2} = \frac{13}{2}
And 
a_3=a_2+d
a_3=\frac{13}{2}+\frac{3}{2} = \frac{16}{2} = 8
Therefore, missing terms are  \frac{13}{2} and 8 
AP series is 5,\frac{13}{2}, 8 , \frac{19}{2}

Posted by

Gautam harsolia

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