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Answers
1. In an AP, it is given that, d = -3, a = 5.
Therefore, nth term is,
Tn = a + (n-1) d
T10 = 5 + (10-1) (-3)
T10 = 5 + (9)(-3)
T10 = 5+(-27)
T10 = -22
Therefore, the tenth term of the AP is -22
2. In the AP, it is given that, a = 3 and d = 5.
Therefore, we have to find n.
Tn = a + (n-1) d
78 = 3 + (n-1) 5
78 - 3 = (n-1) 5
75 = (n-1) × 5
75/5 = (n-1)
15 = n-1
15 + 1 = n
16 = n
Therefore, the 16th term of the AP is 78
3. Numbers divisible by 7 are,
7,14,21,...
For the smallest three digit number,
105/7 = 15
Therefore, a = 105
and for the largest three digit number,
994/7 = 142
Hence, the series,
105, 112, 119,... is an AP.
Now to calculate Tn,
Tn = a + (n-1) d
994 = 105 + (n-1) (7)
994 = 105 + 7n-7
994 - 105 + 7 = 7n
896 = 7n
896/7 = n
128 = n
Therefore, 128 three digit numbers are divisible by 7.
4. In the given AP,
T11 = 38
(a+10d) = 38....(I)
T16 = 73
(a+15d) = 73....(II)
SOLVING (I) AND (II),
a+10d=38
a+15d=73
-5d=-35
5d=35
d=7
Substituting d in (I),
a+10(7)=38
a+70=38
a=38-70
a=-32
Therefore, using the general form of AP,
a,(a+d),(a+2d),(a+3d),....
-32,-25,-18,-11.
Hence, the AP is -32,-25,-18,-11
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