PUMPA - SMART LEARNING
எங்கள் ஆசிரியர்களுடன் 1-ஆன்-1 ஆலோசனை நேரத்தைப் பெறுங்கள். டாப்பர் ஆவதற்கு நாங்கள் பயிற்சி அளிப்போம்
Book Free DemoIf \(b^{th}\), \(i^{th}\), \(s^{th}\) terms of an \(arithmetic\ progression\) are \(x\), \(y\), \(z\) respectively,
then show that \(x(i - s) + y(s - b) + z(b - i) = 0\).
Proof:
Let \(a\) be the first term and \(d\) be the common difference of an \(A.P.\)
\(b^{th} term = x\), That is \(t_b = x\)
\(i^{th} term = y\), That is \(t_i = y\)
\(s^{th} term = z\), That is \(t_s = z\)
Using the general formula:
\(t_b = a + (b - 1)d = x\) - - - - (1)
\(t_i = a + (i - 1)d = y\) - - - - (2)
\(t_s = a + (s - 1)d = z\) - - - - (3)
Now, using the above equations, we get:
\(x(i - s) + y(s - b) + z(b - i)\)
\(= [a + (b - 1)d](i - s) + [a + (i - 1)d](s - b) + [a + (s - 1)d](b - i)\)
\(= [a(i - s) + d(b - 1)(i - s)] + [a(s - b) + d(i - 1)(s - b)] + [a(b - i) + d(s - 1)(b - i)]\)
\(= a[(i - s) + (s - b) + (b - i)] + d[(b - 1)(i - s) + (i - 1)(s - b) + (s - 1)(b - i)]\)
\(= a[i - s + s - b + b - i] + d[bi - bs - i + s + is - ib - s + b+ sb - si - b + i]\)
\(= a(0) + d(0)\)
\(= 0\)
Thus, \(x(i - s) + y(s - b) + z(b - i) = 0\).
Hence, we proved.
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