Question

If the sum of first m terms of an ap is same as the sum of its first n terms, prove that the sum of its m+n terms is 0.

Answer 1

Hi there!

Let a be the first term and d is common difference of the A.P then sum of n terms in A.P

is $S_n$ = (n/2)[ 2a + (n - 1) d]

Similarly um of m terms in A.P is $S_m$ = (m/2)[ 2a + (m - 1) d]

Given that sum of first m terms of an AP is the same as the sum of its first n terms i.e

$S_m$ = $S_n$

⇒  (m/2)[ 2a + (m - 1) d] = (n/2)[ 2a + (n - 1) d]

⇒  (m/2)[ 2a + (m - 1) d] -  (n/2)[ 2a + (n - 1) d] = 0

⇒ 2a[ (m - n)/2)] +[ (1/2)(m²- m - n² + n) d] = 0

⇒ 2a[ (m - n)/2)] +[ (1/2)((m² - n²) - (m - n)) d] = 0

⇒ 2a[ (m - n)/2)] +[ (1/2)((m - n)(m + n) - (m - n)) d] = 0

⇒ (m - n)/2 [2a +[ (m + n) - 1) d] = 0

⇒ [2a +[ (m + n) - 1) d] = 0

∴ 2a = - [ (m + n) - 1) d ----(i)

Sum of first ( m + n ) terms

$S_{m+n}$ = (m+n) / 2 [ 2a + ( m+ n -1) d]

$S_{m+n}$ = (m+n) / 2 [ - [ (m + n) - 1) d + ( m+ n -1) d]            [ from eqn. (i) ]

$S_{m+n}$ = 0.

[ Thank you! for asking the question. ]
Hope it helps!

Answer 2

Hey friend!!


Here's ur answer...


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First term = a
Common difference = d


According to question :

Sum of first m terms = sum of first n terms

$\frac{m}{2} (2a + (m - 1)d) = \frac{n}{2} (2a + (n - 1)d) \\ \\ \frac{m}{2} (2a + md - d) = \frac{m + 1}{2} (2a + md) \\ \\ \frac{2am + {m}^{2}d - md }{2} = \frac{2am + {m}^{2} d + 2a + md}{2} \\ \\ 2am + {m}^{2} d - md = 2am + {m}^{2} d + 2a + md \\ \\ - md = 2a + md \\ - 2md = 2a \\ a = - md \\ \\ \\ now \: taking \: sum \: of \: m \: + n \: \: terms \: \\ \\ = \frac{m + n}{2} (2a + (m + n - 1)d) \\ \\ = \frac{m + m + 1}{2} (2( - md) + (m + m)d) \\ \\ = \frac{2m + 1}{2} ( - 2md + 2md) \\ \\ = \frac{2m + 1}{2} (0) \\ \\ = 0$




Hence proved.



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Hope it may help you....



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