brstorewala
What is the tenth term in a series of 89 consecutive positive integers?
1) The average of the integers is 100
2) The 79th term in the series is 134
Consecutive integers represent arithmetic progression.
For AP:Sequence \(A1, A2, ... An\) so that \(A(n)=A(n-1)+d\) (d constant, common difference).
\(An=A1 + d*(n-1)\) - formula for nth term.
\(Sn=n*\frac{(A1+An)}{2}\) or \(Sn=n*\frac{2*A1+d(n-1)}{2}\) - formula for sum of AP.
In case AP is consecutive integers we'll get common difference, \(d\), as \(1\). So:
\(An=A1+n-1\) - formula for nth term
\(Sn=n*\frac{(A1+An)}{2}\) or \(Sn=n*\frac{2*A1+n-1}{2}\) - formula for sum of AP.
(1) The average of the integers is 100
--> \(Sum=89*100=8900\)
--> \(Sn=n*\frac{2*A1+n-1}{2}=8900\), \(n=89\)
--> \(89*\frac{2*A1+89-1}{2}=8900\).
--> \(A1=56\)
--> \(An=A1+n-1\)
--> \(A10=A1+10-1=56+10-1=65\).
Sufficient.
(2) The 79th term in the series is 134
--> \(A79=134=A1+79-1\)
--> \(A1=56\)
--> \(An=A1+n-1\)
--> \(A10=A1+10-1=56+10-1=65\).
Sufficient.
Answer: D.