If there are 'n' arithmetic means between 'a' and 'b', then d=
A) ab/n+1
B) n+1/b+a
C) b-a/n+1
D) b+a/n+1
Answers
Answered by
36
HELLO DEAR,
LET a1,a2,.......,an BE n ARITHMETIC MEAN
a,a1,a2,.......,an,b ARE in A.PNo OF TERMS
=n+2let d BE THE COMMON DIFFERENCE.
,
b=a+(n+2−1)db−a=(n+1)d⇒d=b−an+1
![b = a + (n + 2 - 1)d \\ = > b - a = (n + 1)d \\ = > d = \frac{(b - a)}{(n + 1)} b = a + (n + 2 - 1)d \\ = > b - a = (n + 1)d \\ = > d = \frac{(b - a)}{(n + 1)}](https://tex.z-dn.net/?f=b+%3D++a+%2B+%28n+%2B+2+-+1%29d+%5C%5C++%3D++%26gt%3B+b+-+a+%3D+%28n+%2B+1%29d+%5C%5C++%3D++%26gt%3B+d+%3D++%5Cfrac%7B%28b+-+a%29%7D%7B%28n+%2B+1%29%7D+)
I HOPE ITS HELP YOU DEAR,
THANKS
LET a1,a2,.......,an BE n ARITHMETIC MEAN
a,a1,a2,.......,an,b ARE in A.PNo OF TERMS
=n+2let d BE THE COMMON DIFFERENCE.
,
b=a+(n+2−1)db−a=(n+1)d⇒d=b−an+1
I HOPE ITS HELP YOU DEAR,
THANKS
Akhilkumar01:
U r cut and pasting the answers
Answered by
5
Answer:
Step-by-step explanation:
1. Let A1, A2,A3,......so on An be 'n' arithmetic means between a and b.
2. Here we see that a, A1,A2,A3,......,b are in AP.
3. Then,
nth term= a+(n-1)d
4. Here, nth term=b
a=a
n= n+2 (because there are total n Arithmetic means and 'a' and 'b'.
5. Now,
b=a+(n+2-1)d
b-a=(n+1)d
b-a/n+1=d
So we got option C.
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