Sum of first hundred numbers common to the two A.P.’s 12, 15, 18, … and 17, 21, 25 …..is
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For AP 1
a=12,d=3,n=100
Sn=n/2(2a+(n-1)d)
S100=100/2(2×12+(100-1)×3)
=50×(24+297)
=16,050
For AP 2
a=17,d=4,n=100
S100=100/2(2×17+(100-1)×4)
=50(28+396)
=21,200
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Answer:
For AP 1
a=12,d=3,n=100
Sn=n/2(2a+(n-1)d)
S100=100/2(2×12+(100-1)×3)
=50×(24+297)
=16,050
For AP 2
a=17,d=4,n=100
S100=100/2(2×17+(100-1)×4)
=50(28+396)
=21,200
Step-by-step explanation:
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