1.
The sum of n terms of an A.P. is 3n2 + n, find the nth term.
• A.
6n - 4
• C.
4n - 4
• B.
6n - 2
• D.
4n - 2
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Explanation :
Given Sn = 3n2 + n, put n = 1, 2
S1 = 3.12 + 1 = 4
T1 = 4
S2 = 3.22 + 2 = 14
T2 = S2 - S1 = 14 - 4 = 10
d = T2 - T1 = 10 - 4 = 6
Tn = a + (n -1) d = 4 + (n - 1) 6 = 6n - 2
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2.
Find the sun of the following series
3 + 7 + 11 + 15 + ����. To 30 terms
• A.
1920
• C.
1970
• B.
1830
• D.
1740
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Explanation :
The given series is 3 + 7 + 11 + 15 + �� to 30 terms.
a = 3, d = 4,� n = 30
We know that
Sn =
 n 2
[2a + (n -1) d]
S30 =
 30 2
[2 x 3 + (30 - 1) (4)]
= 15 [6 + 116]
= 1830
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3.
Find the position of 62 in the following series 2, 5, 8, ....?
• A.
26
• C.
21
• B.
23
• D.
20
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Explanation :
It is an A.P. series with a = 2, d = 3
Let 62 be the nth term.  Tn = 62
But we know that Tn = a + (n � 1) d
62 = 2 + (n � 1) 3
or 62 = 2 + 3n � 3
or 62 � 2 + 3 = 3n
or 63 = 3n
or� 3n = 63
or n = 21
62 is the 21st term
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4.
If you save 1 paise totay, 2 paise next day and 3 paise the succeeding day and so on, what will be your savings in 365 days?
• A.
666.75
• C.
665.35
• B.
668.85
• D.
667.95
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Explanation :
Savings on successive days are 1, 2, 3, 4 �. Paise which form and A.P. with a = 1, d =1
Total savings in 365 days
= S365 =
 365 2
[2 x 1 + (365 - 1) (1)]
=
 365 2
[2 + 364]
=
 365 2
 = 365 x 183
= 66795 paise
= Rs. 667.95
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5.
Find the sun of the following series
72 + 70 + 68 + ����. + 40
• A.
886
• C.
952
• B.
918
• D.
988
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Explanation :
The given series is 72 + 70 + 68 + .... + 40
a = 72, d = -2, l = 40
We know: l = a + (n � 1) d, putting values of a, d and l
40 = 72 + (n � 1) (-2) = 74 � 2n
or 2n = 34
n = 17
Now, we know that: Sn =
 n 2
[a + l]
S17 =
 17 2
(72 + 40) =
 17 2
 = 17 x 56 = 952
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