The $4 t h$ term from the end of the AP
$- 11, - 8, - 5, . . . . . . . . . . . . . . . . . . . .49$ is

37

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40

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43

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58

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Solution

The correct option is B $40$
Given series of AP:
$- 11, - 8, - 5, . . . . . . .49$
New series : $49, . . . . . . . . - 5, - 8, - 11$
In new series of AP first term $a = 49,$ common difference $d = (- 8) - (- 5) = - 3$
$a_{4} = a + 3 d = 49 + 3 (- 3) = 49 - 9 = 40$
$∴$ Option B is correct.

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Similar questions

4^{t h}

The $4^{t h}$ term from the end of an AP is $- 11, - 8, - 5, \dots . ., 49$ is
$A . 37$
$B . 40$
$C . 43$
$D . 58$

4^{t h}

11, 8, 5, . . ., - 49

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