Find five Consecutive terms of an A.P. Such That sum of first three terms is 6 & sum of the remaining two terms is 14 .
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Hey Sis I saw ur answer..u asked my name...it's given in my id...
well let me answer ur question...
Let the terms of AP be
A−2d,a−d,a,a+d,a+2d
A/Q
(a−2d)+(a−d)+a+a(a+d)+(a+2d)=60
5a=60⇒a=12
and a×(a+d)=172+a+2d
⇒a
2
+ad=172+a+2d
⇒144+12d=172+12+2d
⇒12d−2d=184−144=40
⇒10d=40⇒d=4
∴a=12,d=4
and the terms are
12−4×2,12−4,12,12+4,12+2×4
=4,8,12,16,20
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Hello didi..... mai upar wala answer ka malik hu...
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aapne mujse pucha tha ki mai kaisa hu....
Mai thik hu..... Waise "Hum fit toh india fit"
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