If 10 times the 11\(^{th}\) term is equal to 11 times the 14\(^{th}\) term, show that 44\(^{th}\) term is zero.
Proof:
If \(a\) is the first term, \(d\) is the common difference and the \(n^{\text{th}}\) denoted by \(t_n\) can be written as \(t_n = + \).
10 \(\times\) 11\(^{\text{th}}\) term \(=\) 11\( \times\) 14\(^{\text{th}}\) term
10\(\times t_{11}=11\times t_{14}\)
10\( (a + (\)11\( - 1)d) = \)11 \((a + (\)14\( - 1)d)\)
10\( (a + \)\(d) = \)11 \((a + \)\(d)\)
10\(a +\) \(d = \)11\(a + \)\(d\)
\( a + \)\( d = 0\)
\(t_{44}\)\( = 0\)
Hence, proved.