Question
If m≠n and the sequences p,a,b,q and p,m,n,q each are in AP, then b−an−m is ___.
Answer: Option C
:
C
Given, p,a,b,q are in AP.
∴2a=p+b
⇒2a−b=p....(i)
Also,2b=a+q.
⇒2b−a=q....(ii)
Also given that p,m,n,q are in AP.
∴2m=p+n
⇒2m−n=p...(iii)
Also,2n=m+q.
⇒2n−m=q...(iv)
From eqns. (i) & (iii), we get
2a−b=2m−n....(v)
From eqns. (ii) & (iv), we get
2b−a=2n−m....(vi)
Subtracting (vi) from (v), we get
2a−b−(2b−a)=2m−n−(2n−m).
⇒2a−b−2b+a=2m−n−2n+m
3a−3b=3m−3n
b−a=n−m
⇒b−an−m=1
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:
C
Given, p,a,b,q are in AP.
∴2a=p+b
⇒2a−b=p....(i)
Also,2b=a+q.
⇒2b−a=q....(ii)
Also given that p,m,n,q are in AP.
∴2m=p+n
⇒2m−n=p...(iii)
Also,2n=m+q.
⇒2n−m=q...(iv)
From eqns. (i) & (iii), we get
2a−b=2m−n....(v)
From eqns. (ii) & (iv), we get
2b−a=2n−m....(vi)
Subtracting (vi) from (v), we get
2a−b−(2b−a)=2m−n−(2n−m).
⇒2a−b−2b+a=2m−n−2n+m
3a−3b=3m−3n
b−a=n−m
⇒b−an−m=1
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