Veto q&a 7 Month22 (1)

Q
nn ( + 1)
given by =
2
×
= 100 101 = 50 x 101
2
101 is an odd number and 50 is divisible by 2.
Hence, 50 x 101 will be divisible by 2
The sum of third and ninth term of an AP is 8. Find
the sum of the first 11 terms of the progression
a) 44 b) 22 c) 19 d) none
ans: a) 44
The 3 rd term t 3
= a + 2d
The 9 th term t 9
= a + 8d
t 3
+ t 9
= 2a + 10d = 8
Sum of first 11 terms of an AP is given by
11
Sn = ( 2a×
10d)
2
11
Sn = × 8 = 44
2
Q Find the AP whose 10 th term is 5 and 18 th term is
77
a) -77, -67 b) -67, -57
c) 76, 67 d) none
ans: a) -77, -67
Given, 10 th term of an AP = 5
a + (n-1)d = 5
a+(10-1)d = 5
⇒a+9d = 5 - - - - - - (1)
and 18 th term = 77
a + (n-1)d = 77
a+(18-1)d = 77
⇒a+17d = 77 - - - - - - (2)
(2) - (1), 8d = 72
∴ d = 9
Substituting the value of d = 9 in (1)
a + 81 - 5, a = 5-81 = -76
∴ The AP is -76, -67
Q The sum of n terms of an AP is 3n2 + n. Find
the n th term
a) 6n-4 b) 6n-2 c) 4n-4 d) 4n-2
ans: b) 6n-2
Q&A Bonus
@hang out
Q
Q
Given Sn = 3n 2 +n, Put n = 1, 2
S 1
= 3x1 2 + 1 = 4
T 1
= 4
S 2
= 3 x 2 2 + 2 = 14
T 2
= S 2
- S 1
=14 - 4 = 10
∴ d = T 2
- T 1
= 10 - 4 = 6
∴ Tn = a + (n-1)d = 4+ (n-1)6 = 6n-2
Find the sum of the following series
72 + 70 + 68 + ........... + 40
a) 886 b) 918 c) 952 d) 988
ans: c) 952
2018
April
The gives series is 72 + 70 + 68 + ............ +
40
a = 72, d = -2, n th term = 40
We know that n th term = a+(n-1)d
Putting values of a, d, and n th term
40 = 72 + (n-1)(-2) = 74 -2n or
2n = 34
∴ n = 17
n
Now we know that : Sn = ( a + 1)
2
17 17
S17
= ( 72 + 40) = ( 112)
2 2
= 17 x 56 = 952
Find the sum of lthe following series
3 + 7 + 11 + 15 + ...... to 30 terms
a) 1920 b) 1830
c) 1970 d) 1740
ans: b) 1830
a= 3, d = 4, n = 30
n
We know that Sn = 2 a+ ( n−1)
d
2
S 30
30
= 2 × 3 + (30 − 1)(4)
2
( )
= 15 (6+116) = 1830
( )
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