question no. 2,7 and 3
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first no divisible by 5 bet.100 and 1000 is 105 and last no. is 995
so
a+(n-1)d=995. (a=first term d=common difference)
105+5n-5=995
5n=895
n=179
sum=179÷2×(105+995)
=179÷2×(1100)
=179÷2×1100
=179×550
=98450
7ans
k th term is 5k+1
means
1st term = 6
2nd=11
3rd = 16
a=6
c.d=5
Sn=n/2(12+(n-1)5)
first no divisible by 5 bet.100 and 1000 is 105 and last no. is 995
so
a+(n-1)d=995. (a=first term d=common difference)
105+5n-5=995
5n=895
n=179
sum=179÷2×(105+995)
=179÷2×(1100)
=179÷2×1100
=179×550
=98450
7ans
k th term is 5k+1
means
1st term = 6
2nd=11
3rd = 16
a=6
c.d=5
Sn=n/2(12+(n-1)5)
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