Question

The sum of first 'm' terms of an A.P is 'n' and the sum of first 'n' terms is 'm', then show that the sum of first (m+n) terms is -(m+n). The most appropriate and fastest answer will be marked as the brainliest at once. Please respond ASAP.

Answer 1

sum of p terms is given by the formula
Sp= p/2=2a +(p-1)d}, where "a" is the first term and "d" is the common difference.

n = m/2{2a+(m-1)d }

m = n/2{2a + (n-1)d}

n-m = m/2{2a+(m-1)d } - n/2{2a + (n-1)d}

n-m = am +m2d/2 - md/2 -an - n2d/2 + nd/2

n-m = a(m-n) + d/2 (m2 - m -n2 +n)

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

n-m = a(m-n) + d/2[(m-n)(m+n-1)]

n-m = (m-n) [a + d/2(m+n-1)]

-(m-n) = (m-n) [a + d/2(m+n-1)]

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

-2 = 2a+d(m+n-1)

since S(m+n)=m+n/2 * [2a+d(m+n-1)]

S(m+n) =m+n/2 * -2

S(m+n) = -(m+n)

I hope this will help you

Answer 2

Note: We know that Sum of first n terms of an AP sn = (n/2)[2a + (n - 1) * d]

(i)

Given : Sum of first 'm' terms of an AP is 'n'.

Sm = n

⇒ n =  (n/2)[2a + (n - 1) * d]

⇒ 2n = m[2a + (m - 1) * d]

⇒ 2n = 2am + m(m - 1) * d

⇒ 2n = 2am + (m^2 - m) * d

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(ii)

Given : Sum of first 'n' terms is 'm'.

Sn = m

⇒ m = (n/2)[2a + (n - 1) * d]

⇒ 2m = 2an + n(n - 1) * d

⇒ 2m = 2an + (n^2 - n) * d

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On solving (1) - (2), we get

⇒ 2n - 2m = 2am - 2an + d(m^2 - m - n^2 + n)

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

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

Divide the equation by m - n on both sides, we get

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

Given Sum of first (m + n) terms:

Sm + n:

⇒ (m + n)/2[2a + (m + n - 1) * d]

⇒ (m + n)/2[-2] {from (1)]

⇒ (m + n)[-1]

⇒ -(m + n).

Therefore, Sum of first (m + n) terms is -(m + n).

Hope this helps!