Question

The 11th term of an A.P. exceeds its 4th term by 14. Find the common difference.
(class 10 CBSE SAMPLE PAPER 2017-18 MATHS)

Answer 1

Let ‘a’ be the first term  and ‘d’  be the  common difference of the given A.P


According to Question


a11= a4+14


a11– a4 =14


[a+(11-1)d] - [a+ (4-1)d= 14


{ an = a+(n-1)d}



a + 10d – (a + 3d) = 14


a + 10d - a - 3d = 14


a-a+10d-3d=14


7d = 14


d=2


Hence, the common difference of this AP is 2.


HOPE THIS WILL HELP YOU....

Answer 2

let first term be a and common difference be d

now according to question
a+10d-(a+3d)=14
a+10d-a-3d=14
7d=14
d=2 ans

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