12. The common difference of an AP is 2.5. By how much does the 50th term of the
AP exceeds its 35th term ?
Answers
Answer:
d = 2.5
Asked :: (a+49d) - ( a+34d) == 15d == 15(2.5) = 37.5
Given : The common difference of an AP is 2.5.
To Find : By how much does the 50th term of the AP exceeds its 35th term
Solution:
AP is
a , a + d , a + 2d
a = first term
d = common difference
aₙ = a + (n-1)d
a₅₀ = a + (50 - 1)d = a + 49d
a₃₅ = a + (35 - 1)d = a + 34d
50th term of the AP exceeds its 35th term
a₅₀ - a₃₅
= a + 49d - ( a + 34d)
= 49d - 34d
= 15d
d = 2.5 given
= 15 (2.5)
= 37.5
of an AP is 2.5. By how much does the 50th term of the AP exceeds its 35th term by 37.5
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