79
The $4^{th}$ and $10^{th}$ term of an A.P. are $13$ and $25$ respectively. Find its $24^{th}$ term.
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$a + 3d = 13$ (i) (1/2 Mark)
$a + 9d = 25$ (ii) (1/2 Mark)
solving (i) & (ii) to get $d = 2, a = 7$ (1 Mark)
therefore $a_{24} = 53$ (1 Mark)
$a + 9d = 25$ (ii) (1/2 Mark)
solving (i) & (ii) to get $d = 2, a = 7$ (1 Mark)
therefore $a_{24} = 53$ (1 Mark)