If the sequence is an A-P- show that a m - n - a m - n -2 a m

It is given that the sequence lt;a_n gt; is an A.P. ∴ a_n = a+(n 1)d …… (i) Similarly from (i) a_m+n =a+(m+n 1)d …… (ii) a_m n =a+(m n 1)d …… (iii) Adding (ii ...

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Standard XII

Mathematics

Sequence

If the sequen...

Question

If the sequence $< a_{n} >$ is an A.P., show that $a_{m + n} + a_{m - n} = 2 a_{m}$

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Solution

It is given that the sequence $< a_{n} >$ is an A.P.

$∴ a_{n} = a + (n - 1) d \dots \dots (i)$

Similarly from (i)

$a_{m + n} = a + (m + n - 1) d \dots \dots (i i)$

$a_{m - n} = a + (m - n - 1) d \dots \dots (i i i)$

Adding (ii) and (iii)

$a_{m + n} + a_{m - n} = (a + (m + n - 1) d) + (a + (m - n - 1) d)$

$= 2 a + ((m + n - 1 + m - n - 1) d)$

$= 2 a + 2 d (m - 1)$

$= 2 (a + (m - 1) d)$

$= 2 a_{m}$

Hence proved.

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

It is given that the sequence <an> is an A.P. ∴an=a+(n−1)d……(i) Similarly from (i) am+n=a+(m+n−1)d……(ii) am−n=a+(m−n−1)d……(iii) Adding (ii) and (iii) am+n+am−n=(a+(m+n−1)d)+(a+(m−n−1)d) =2a+((m+n−1+m−n−1)d) =2a+2d(m−1) =2(a+(m−1)d) =2am Hence proved.

If the sequence $< a_{n} >$ is an A.P., show that $a_{m + n} + a_{m - n} = 2 a_{m}$