IN And AP, THE SUM OF 1st TEN TERNS 210
AND THE Some of THE NEXT TEN TERMS is 610.
FIND AP.
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given s10 =210
s20=610
;s10=n/2[2a+(n-1) d]
210=10/2[2a+(10-1) d]
210=5[2a+9d]
210=10a+45d
42=2a+9d .... (1)
S20=n/2[2a+(n-1) d]
610=20/2[2a+(20-1) d]
610=10[2a+19d]
610=20a+190d
61=2a+19d .... (2)
from (1) &(2)
10d=19
d=19/10
d=1.9 [put in eqn(1) ]
42=2a+9(1.9)
a=17.45
a2=17.45+1.9=19.35
a3=19.35+1.9=21.25
a4=21.25+1.9=23.15
Therefore, an AP is 17.45,19.35,21.25,23.15........
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