An arithmetic sequence has as its 5th term equal to $22$ and its 15th term equal to $62$ . Find its 100th term.

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Solution

Given : $t_{5} = 22$ and $t_{15} = 62$
We know, $t_{n} = a + (n - 1) d$
So, $t_{5} = a + (5 - 1) d \Rightarrow a + 4 d = 22$ ....(I)
And, $t_{15} = a + (15 - 1) d \Rightarrow a + 14 d = 62$ ....(II)
Subtract eq.(I) from eq. (II), we get
$\Rightarrow 10 d = 40$
$\Rightarrow d = 4$
Hence, $a = 6$
So, $t_{100} = 6 + (100 - 1) 4$
$∴ t_{100} = 402$

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