Question

please answer this question my dear brainly friends​

Answer 1

Answer:

Proved

Step-by-step explanation:

Let the first term if AP is a,common difference d

Sum of first n terms

$S_n = \frac{n}{2} (2a + (n - 1)d) = p \: (let) \\ \\ 2an + {n}^{2}d - nd = 2p \: \: ...eq1 \\ \\ S_m = \frac{m}{2} (2a + (m - 1)d) = p \: \\ \\ 2am + {m}^{2}d - md = 2p \: \: ...eq2 \\ \\ subtract \: eq1 \: from \: \: eq2 \\ \\ 2am + {m}^{2}d - md - 2an - {n}^{2}d + nd = 2p - 2p \\ \\ 2a(m + n) + d( {m}^{2} - {n}^{2} - (m - n)) = 0 \\ \\ (m + n)(2a + (m + n - 1)d) = 0 \\ \\ (2a + (m + n - 1)d) = 0 \: \: \: ..eq3\\ \\ sum \: of \: (m + n)terms \\ \\ S_{m + n} = \frac{m + n}{2} (2a + (m + n - 1)d) \\ \\ substitute \: value \: from \: eq3 \\ \\ S_{m + n} = \frac{m + n}{2} \times 0 \\ \\ S_{m + n} = 0$

Proved

Hope it helps you.