The pth, qth and rth terms …
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Sia ? 6 years, 6 months ago
Let the first term be a and the common difference be d.
{tex}a = a' + (p - 1)d, b = a' + (q - 1)d, c = a' + (r - 1)d{/tex}
{tex}a(q-r) = [a' + (p - 1)d] [(q - r)]{/tex}
{tex}b(r - p) = [a' + (q - 1)d][r - p]{/tex}
and {tex}c[p - q ] = [a' + (r - 1)d][p - q]{/tex}
{tex}\therefore{/tex} {tex}a(q - r) + b(r - p) + c(p - q) = [a' + (p - 1)d] [(q - r)] + [a' + (q - 1)d][r - p] + [a' + (r - 1)d][p - q]{/tex}
{tex}= a' [q - r + r - p + p - q] + d [p (q - r)-q + r + (q - 1) (r - p) + (r -1) (p - q)]{/tex}
{tex}= a'\times0 + d[ pq - pr + qr - pq + pr - qr + (-q + r - r + p - p + q)]{/tex}
= 0
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