Question:
If the sequence < an > is an A.P., show that
am +n +am − n = 2am.
Solution:
Let the sequence < an > be an A.P. with the first term being A and the common difference being D.
To prove: am +n +am − n = 2am
LHS: am +n +am − n
$=A+(m+n-1) D+A+(m-n-1) D \quad\left\{\because a_{n}=a+(n-1) d\right\}$
$=A+m D+n D-D+A+m D-n D-D$
$=2 A+2 m D-2 D \quad \ldots(\mathrm{i})$
RHS: 2am
$=2[A+(m-1) D]$
$=2 A+2 m D-2 D \quad \ldots(\mathrm{ii})$
From (i) and (ii), we get:
LHS = RHS
Hence, proved.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.