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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
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The number of terms in the AP a, b,....c is

$(a)\;c\qquad(b)\;\frac{(c-a)}{(b-a)}\qquad(c)\;\frac{(b+c-2a)}{(b-a)}\qquad(d)\;\frac{(b+c+2a)}{(b-a)}$

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1 Answer

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Answer : (c) $\;\frac{(b+c-2a)}{(b-a)}$
Explanation : Let d be the common differance of the AP
Then d=b-a
The value of c is given by
$c=a+(n-1)d$
$c=a+(n-1)(b-a)$
Solving this , $ c=2a-b\;+(b-a)n$
$n=\frac{(b+c-2a)}{(b-a)}.$
answered Jan 17, 2014 by yamini.v
 

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