The 11 term of ap exceeds its 4 term by 14 find the common difference
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I think the answer is this Sir
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hello,
let the first term be a and common difference d
11th term is given by a+(n-1)d= a+(11-1)d
=a+10d
4th term is given by a+(n-1)d= a+(4-1)d
=a+3d
now it is given that 11th term exceeds 4th term by 14
i.e. 11th term - 4th term =14
a+10d-(a+3d)=14
a+10d-a-3d=14
=7d=14
d=2
therefore the common difference is 2
hope this helps please mark it as brainliest if u like it
let the first term be a and common difference d
11th term is given by a+(n-1)d= a+(11-1)d
=a+10d
4th term is given by a+(n-1)d= a+(4-1)d
=a+3d
now it is given that 11th term exceeds 4th term by 14
i.e. 11th term - 4th term =14
a+10d-(a+3d)=14
a+10d-a-3d=14
=7d=14
d=2
therefore the common difference is 2
hope this helps please mark it as brainliest if u like it
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