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If the Sequence < an > is an A.P., Show that Am +N +Am − N = 2am. - Mathematics

If the sequence < an > is an A.P., show that am +n +am − n = 2am.

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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

\[= A + (m + n - 1)D + A + (m - n - 1)D { \because a_n = a + (n - 1)d}\]

\[ = A + mD + nD - D + A + mD - nD - D\]

\[ = 2A + 2mD - 2D . . . (i)\]

RHS: 2am

\[= 2[A + (m - 1)D]\]

\[ = 2A + 2mD - 2D . . . (ii)\]

From (i) and (ii), we get:
LHS = RHS
Hence, proved.

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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 19 Arithmetic Progression
Exercise 19.2 | Q 2 | Page 12
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