Question

Plzz answer as soon as possible....n....no spams please....​

Answer 1

Step-by-step explanation:

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Answer 2

Given:

Sum of ‘m’ terms = Sum of ‘n’ terms

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

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

(m)*(2a + (m - 1)*d) - (n)*(2a + (n - 1)*d) = 0

2am + m(m - 1)d - 2an - n(n - 1)d = 0

2a(m - n) + m^2*d - m*d - n^2*d + n*d = 0

2a(m - n) + m^2*d - n^2*d - m*d + n*d = 0

2a(m - n) + (m^2 - n^2)*d - (m - n)*d = 0

2a(m - n) + (m + n)*(m - n)*d - (m - n)*d = 0


(m - n)[2a + (m + n)*d - d] = 0

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

Multiplying (m + n)/2 on both sides,

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

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

S(m +n) = 0

Hence Proved