Question

12. The common difference of an AP is 2.5. By how much does the 50th term of the
AP exceeds its 35th term ?
​

Answer 1

Answer:

d = 2.5

Asked :: (a+49d) - ( a+34d) == 15d == 15(2.5) = 37.5

Answer 2

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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