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Q1.
Find the minimum value of 9x + 91 - x , x ∈ R
(a) 6
(b) 8
(c) 5
(d) 9
Solution. a
As we know that
AM ≥ GM
9x + 91 - x/2
≥ √( 9
x + 9
1 - x)
9
x + 9
1 - x ≥ 2√(9)
9
x + 9
1 - x ≥ 6
So the minimum value of 9
x + 9
1 - x is 6 .
Solution
Q2.
If a, b, c are in A.P. then the value of a - 2b + c is
(a) 2
(b) 1
(c) 0
(d) 5
Solution. c
2b = a + c
= (a + c) - 2b = 2b - 2b = 0
Solution
Q3.
If T10 = 30, find S19
(a) 570
(b) 550
(c) 500
(d) 650
Solution. a
T
10 = a + 9d = 30
S
19 =
19/2
[2a + 18d]
S
19 = 19 x [a + 9d] =
19 x (30)
= 570
Solution
Q4.
If the sum of first n terms of an A.P. is 3n2 + 4n, then find the 10th term.
(a) 55
(b) 65
(c) 60
(d) 61
Solution. d
Sn = 3n2 + 4n
If we put n = 1, we get first term
S1 = a = 7
and common
difference (d) = 2x Coefficient of n2 = 6
Tn = a + (n - 1)d
T10 =
7 + (9)6
= 7 + 54
= 61
Solution
Q5.
An A.P. has 13 terms whose sum is 143 . The third term is 5. So the first term is .
(a) 2
(b) 1
(c) 3
(d) 0
Solution. a
As per question
13/2
[2a + 12d] = 143
a + 6d = 11 --------------(1)
a + 2d = 5 --------------- (2)
Solving both equation
4d = 6
d =
3/2
a + 2d = 5
a + 2(
3/2
) = 5
a + 3 = 5
a = 2
Solution
