11th term of the A-P -3- -1 2 -2- is

a= 3 d= 12 ( 3)= 12+3=52 T_11=a+10d =( 3)+10(52) = 3+25 =22 . ...

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Byju's Answer

Standard VI

Mathematics

Divisibility Tests for 4

11th term of ...

Question

$11 t h$ term of the $A . P - 3, \frac{- 1}{2}, 2, . . .,$ is

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Solution

$a = - 3$
$d = \frac{- 1}{2} - (- 3) = \frac{- 1}{2} + 3 = \frac{5}{2}$
$T_{11} = a + 10 d$
$= (- 3) + 10 (\frac{5}{2})$
$= - 3 + 25$
$= 22$ .

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Similar questions

Q. Find the

11^{t h}

term of the A.P :

- 3, - \frac{1}{2}, 2....

Which of the following is the

11 t h

term of the A.P. :

- 3, - \frac{1}{2}, 2, . . .

11^{t h}

term of the AP:

- 3,

- \frac{1}{2}, 2, . . .

Choose the correct choice in the following and justify:
(ii) $11^{t h} t e r m o f t h e A . P - 3, \frac{- 1}{2}, 2 i s$

A) 28
B) 22
C) -38
D) 44

Question 2 (ii)
Choose the correct choice in the following and justify.
(ii) $11^{t h} t e r m o f t h e A . P - 3, \frac{- 1}{2}, 2 i s$

A) 28
B) 22
C) -38
D) 44

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11th term of the A.P−3,−12,2,..., is

a=−3d=−12−(−3)=−12+3=52T11=a+10d=(−3)+10(52)=−3+25=22.

$11 t h$ term of the $A . P - 3, \frac{- 1}{2}, 2, . . .,$ is

$a = - 3$
$d = \frac{- 1}{2} - (- 3) = \frac{- 1}{2} + 3 = \frac{5}{2}$
$T_{11} = a + 10 d$
$= (- 3) + 10 (\frac{5}{2})$
$= - 3 + 25$
$= 22$ .