Two arithmetic progressions have the same common difference. Their first terms are A and B respectively. The difference between their nth terms is
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Answer:
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Given two A.P have same common difference =d
let the first term of first A.P be a
1
and the first term of second A.P be a
1
′
hence 100th term of first A.P is given by
a
100
=a
1
+(100−1)d
⟹a
100
=a
1
+99d...eq(1)
and 100th term of second A.P is given by
a
100
′
=a,
1
+(100−1)d
⟹a
100
′
=a
1
′
+99d...eq(2)
and given that the difference between their 100th term is 100
hence, a
100
−a
100
′
=100
⟹(a
1
+99d)−(a
1
′
+99d)=100
⟹a
1
−a
1
′
=100...….eq(3)
1000th term of first A.P is
a
1000
=a
1
=(1000−1)d
⟹a
1000
=a
1
+999d...….eq(4)
and 1000th term of second A.P is
a
1000
′
=a
1
′
=(1000−1)d
⟹a
1000
′
=a
1
′
+999d...….eq(4)
now difference of their 1000th term is given by
a
1000
−a
1000
′
=(a
1
+999d)−(a
1
′
+999d)
a
1000
−a
1000
′
=a
1
−a
1
′
put value of a
1
−a
1
′
=100 from eq(3) in above equation we get
a
1000
−a
1000
′
=100
hence the difference between their 1000th term is 100.