If $t_{54}$ of an A.P. is $- 61$ and $t_{4} = 64,$ find $t_{10} .$

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Solution

Let $a$ be the first term and $d$ be the common difference
so $t_{54} = a + 53 d = - 61$ (i)
and $t_{4} = a + 3 d = 64$ (ii)
equation (i) $-$ (ii),
$\Rightarrow$ $50 d = - 125$
$⟹ d = - \frac{5}{2}$
On putting value of $d$ in equation (ii)
$⟹ a - \frac{3 \times 5}{2} = 64$
$⟹ a = 64 + \frac{15}{2}$
$\Rightarrow$ $a = \frac{143}{2}$
so, $t_{10} = a + 9 d$
$⟹ t_{10} = \frac{143}{2} + 9 (- \frac{5}{2}) = 49$

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