Find the $6^{t h}$ term from the end of the A.P $17, 14, 11, . . . - 10$

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Solution

$17, 14, 11, . . . - 10$ are in A.P

or $- 10, - 7, - 4, . . .17$ are in A.P

$a = - 10$ and $d = - 7 - (- 10) = - 7 + 10 = 3$

$∴ t_{6} = a + 5 d = - 10 + 5 \times 3 = - 10 + 15 = 5$

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