Find $T_{20} - T_{10}$ for the A.P. $- 3, - 5, - 7, - 9, . . . . . . . . . . .$

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Solution

The correct option is C $- 20$
Given series is $- 3, - 5, - 7, - 9, . . . .$
Here, first term $a$ $=$ $- 3$ and common difference $d$ $=$ $- 5 - (- 3)$ $=$ $- 5 + 3 = - 2$
Now, $T_{n} = a + (n - 1) d$
Therefore, $T_{20} = a + (20 - 1) d = a + 19 d$ and
$T_{10} = a + (10 - 1) d = a + 9 d$
Now, $T_{20} - T_{10} = (a + 19 d) - (a + 9 d)$
$= 10 d = 10 (- 2) = - 20$

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Similar questions

Class:	$3 - 6$	$6 - 9$	$9 - 12$	$12 - 15$	$15 - 18$	$18 - 21$	$21 - 24$
Frequency:	$2$	$5$	$10$	$23$	$21$	$12$	$3$

T_{25} - T_{20} = 15

∴

d =

40

5, 3, 10, 20, 25, 11, 13, 7, 12, 31, 19, 10, 12, 17, 18, 11, 32, 17, 16, 2, 7, 9, 7, 8,

3, 5, 12, 15, 18, 3, 12, 14, 2, 9, 6, 15, 15, 7, 6, 12

14

26

Number	$0 - 3$	$3 - 6$	$6 - 9$	$9 - 12$	$12 -$	$15 - 18$	$18 - 19$	$21 -$
Frequency	$4$		$9$	$7$	$6$	$3$	$1$

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