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The sum of an AP is 1815. If the first and last terms are -12 and 133 respectively, find the number of terms in the AP
$(a)\;15\qquad(b)\;133\qquad(c)\;30\qquad(d)\;12$
jeemain
math
class11
unit7
sequences-and-series
medium
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asked
Nov 19, 2013
by
harini.tutor
edited
Apr 1, 2016
by
pady_1
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1 Answer
Toolbox:
In an AP, $ S_{n} = \large \frac {n} {2} (a + l) $
In an AP, $ S_{n} = \large \frac {n} {2} (a + l) $
$ \therefore 1815 = \large \frac {n} {2} (-12 + 133) $
$ 121n = 1815 X 2 $
n = 30
answered
Nov 19, 2013
by
harini.tutor
edited
Nov 19, 2013
by
harini.tutor
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