2. (i)Which term of the A.P. 3,8,13,...is 248?
(ii) Which term of the A.P. 84,80,76,... is 0?
(iii) Which term of the A.P. 4,9,14,... is 254?
(iv) Which term of the A.P. 21,42,63,84,...is 420?
(v) Which term of the A.P. 121,117,113,... is its first negative term?

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Solution

1.
$a = 3, d = 8 - 3 = 5, a_{n} = 248, n = ? a_{n} = a + (n - 1) d 248 = 3 + (n - 1) \times 5 248 - 3 = (n - 1) \times 5 245 = (n - 1) \times 5 245 \div 5 = n - 1 49 = n - 1 n = 49 + 1 n = 50 2 - W e k n o w t h a t T_{n} = a + (n - 1) d = 84 + (n - 1) \times (- 4) = 0 = 84 - 4 n + 4 = 0 = 88 - 4 n = 0 n = 88 / 4 = 22 3 - H e r e t h e f i r s t t e r m = a = 4 a n d t h e c o m m o n d i f f e r e n c e = d = 9 - 4 = 5 L e t n^{t h} t e r m o f t h e g i v e n A . P . b e 254, t h e n,$
$a \frac{}{n} = 254 \\ \\ = > a + (n + 1)d = 254 \\ \\ = > 4 + (n - 1) 5 = 254 \\ \\ = > 5n = 255 \\ \\ = > n = 51$

Hence, the 51 th term of the given A.P. is 254
4-
$a = 21 d = 21 a_{n} = 420 n = ?$
$a_{n}$ =a+(n-1)d
420=21+(n-1)21
(n-1) x 21=399
n-1=399/21
n-1=19
n=19+1=20
ans..therefore the 20th term of the AP is 420
5-
121,117,113.......
a=121. d=-4
For a(n) be negative
a(n)<0
a+(n-1)d <0
121+(n-1) x (4)<0
For ans be negative
(n-1)(-4)>121
4n >(125)
n >125/4
n >31.25
n=32
Therefore the 31st term is the negative term
Verification
a(n)=a+(n-1)d
a(n)=121+(32-1)-4
a(n)=121+31 x -4
a(n)=121-124
a (n)=-3

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(vi) Which term of the A.P. –7, –12, –17, –22,... will be –82? Is –100 any term of the A.P?

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