using AP, find the sum of all 3-digit natural numbers which are the multiples of 7.
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Three digit numbers which are divisible by 7 are :105,112,119---------994
a1=105,
a2=112
a3=119
an=994
d=a2-a1
d=112-105=7
d=7
an=a+(n-1)d
994=105+(n-1)7
994-105=(n-1)7
889=(n-1)7
889/7=(n-1)
127=n-1
127+1=n
128 =n
128 three digit numbers are divisible by 7
i hope thus helps uuuu
a1=105,
a2=112
a3=119
an=994
d=a2-a1
d=112-105=7
d=7
an=a+(n-1)d
994=105+(n-1)7
994-105=(n-1)7
889=(n-1)7
889/7=(n-1)
127=n-1
127+1=n
128 =n
128 three digit numbers are divisible by 7
i hope thus helps uuuu
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Answer:
128 is the no. of 3 digit numbers which are divisible by 7
Step-by-step explanation:
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