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Show that the sum of (*m* + *n*)^{th} and (*m* – *n*)^{th} terms of an A.P. is equal to twice the *m*^{t}^{h} term.

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#### Solution

Let *a* and *d* be the first term and the common difference of the A.P. respectively.

It is known that the *k*^{th} term of an A. P. is given by

*a*_{k} = *a* + (*k* –1) *d*

∴ *a*_{m}_{ + }_{n} = *a* + (*m* + *n* –1) *d*

*a*_{m}_{ – }_{n} = *a* + (*m* – *n* –1) *d*

*a*_{m}_{ }= *a* + (*m* –1) *d*

∴ *a*_{m}_{ + }_{n} + *a*_{m}_{ – }* _{n}* =

*a*+ (

*m*+

*n*–1)

*d*+

*a*+ (

*m*–

*n*–1)

*d*

= 2*a* + (*m* + *n* –1 + *m* – *n* –1) *d*

= 2*a* + (2*m* – 2) *d*

= 2*a* + 2 (*m* – 1) *d*

*=*2 [*a* + (*m* – 1) *d*]

= 2*a*_{m}

Thus, the sum of (*m* + *n*)^{th} and (*m* – *n*)^{th} terms of an A.P. is equal to twice the *m*^{t}^{h} term.

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