find the sum of first 12multiples of2 and3 using the formula
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For multiple of 2.
Sn=12,a=2,d=2
Sn=n/2{2a+(n-1)d}
S12= 12/2{2×2+(12-1)2}
=6{4+11×2}
=6{4+22}
=6×88
=528
there for sum of first 12 multiple of 2 is 528
for multiple of 3
Sn=12,a3,d=3
Sn=n/2{2a+(n-1)d}
S12= 12/2{2×3+(12-1)3}
= 6{6+11×3}
= 6{6+33}
= 6×198
= 98
there for sum of first 12multiple of 3 is 988
Sn=12,a=2,d=2
Sn=n/2{2a+(n-1)d}
S12= 12/2{2×2+(12-1)2}
=6{4+11×2}
=6{4+22}
=6×88
=528
there for sum of first 12 multiple of 2 is 528
for multiple of 3
Sn=12,a3,d=3
Sn=n/2{2a+(n-1)d}
S12= 12/2{2×3+(12-1)3}
= 6{6+11×3}
= 6{6+33}
= 6×198
= 98
there for sum of first 12multiple of 3 is 988
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