please solve question no. 14
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consider a and d as the first term and the common difference of an AP respectively.
nth term of an AP,an=a+(n-1)d
given that
s10=n/2{2a+(n-1)d}
=10/2{2a+9d}=210
=2a+9d=42(1)
15th term from the last=(50-15+1)the=36 from the beginning 36
a36=a+35d
s15=15/2 {2a36+(15-1)d}=256
=15/2{2(a+35d)+14d}=2565
=15(a+35d+7d)=2565
a+42d=171(2)
from (1)and (2),we have d=4 and a=3
therefore the terms in AP are 3,7,11...and 199.
hope this will help you
nth term of an AP,an=a+(n-1)d
given that
s10=n/2{2a+(n-1)d}
=10/2{2a+9d}=210
=2a+9d=42(1)
15th term from the last=(50-15+1)the=36 from the beginning 36
a36=a+35d
s15=15/2 {2a36+(15-1)d}=256
=15/2{2(a+35d)+14d}=2565
=15(a+35d+7d)=2565
a+42d=171(2)
from (1)and (2),we have d=4 and a=3
therefore the terms in AP are 3,7,11...and 199.
hope this will help you
shalu3211:
your welcome
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