Question

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

Answer 1

Given,

Common difference of AP = 2.5

To find,

The difference between the 35th term and 50th term.

Solution,

We can easily solve this mathematical problem by using the following mathematical process.

Now, each term of AP exceeds by the value of the common difference, than it's previous term.

Number position difference between the two terms

= 50 - 35

= 15

Value difference between the two terms

= Number position difference × Common difference

= 15 × 2.5

= 37.5

Hence, the 50th term of AP exceeds by 37.5 than it's 35th term.

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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