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The number of terms in an A.P. is even ; the sum of the odd terms in it is 24 and that the even terms is 30 . If the last term exceeds the first term by $\;10 \large\frac{1}{2}\;$, then the number of terms in the A.P. is :

$(a)\;4\qquad(b)\;8\qquad(c)\;12\qquad(d)\;16$

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Let there be n number of terms.
Out of which n/2 are even or odd.
Let first term be a  , last be a+(n-1)d.
Given

(n-1)d=21/2  

And, a+(a+2d)+(a+4d)+....a+(n-2)d=24     (n/2 terms)
a+d+(a+3d)+(a+5d)+.....+a+(n-1)d=30      (n/2 terms)
Subtracting we get , nd/2=6
or nd=12

Solving both equation gives , n=8
answered Apr 2, 2015 by Limitsofawesomeness
 

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