1
UDAAN - 2023
ARITHMETIC PROGRESSIONS
1.
6th term from the end of the A.P. 17, 14, 11,., (–40)
is
(A) –20
(B) 15
(C) –25
(D) 30
2.
The sum of all even numbers between 5 and 100 is
(A) 2444
(B) 2400
(C) 2445
(D) 2450
3.
Number of terms of the A.P. 21, 18, 15, that must be
added to get the sum equal to zero is
(A) 12
(B) 13
(C) 14
(D) 15
4.
In an A.P., the first term is 22 and nth term is –11. If
the sum to first n terms is 66, then the value of n is
(A) 10
(B) 12
(C) 14
(D) 16
5.
A sum of Rs. 1200 is invested at 5% simple interest
per annum. Calculate the interest at the end of 1, 2, 3,
...... years. Is the sequence of interests an A.P.? Find
the interest at the end of 20 years.
(A) 1200
(B) 1400
(C) 1600
(D) 1800
6.
If 280 logs are stacked in such a way that there are
25 logs in the bottom row, 24 in the next row, 23 in
the next row and so on, then the number of rows in
which 280 logs are placed is
(A) 15
(C) 17
DHA-02
(B) 16
(D) 18
7.
A sum of Rs. 800 is to be used to give ten cash prizes
to students of a school for their overall academic
performance. If each prizes is Rs. 10 less than its
preceding prize, then the value of the first prize is
(A) Rs. 115
(B) Rs. 125
(C) Rs. 135
(D) Rs. 145
8.
If xth term of A.P. is
term is
(A) 1
(C)
9.
1
x
1
1
and yth term is
then (xy)th
y
x
(B) 0
(D) None of these
Two APs have the same common difference. The first
term of one of these is –1 and that of the other is –8.
Then the difference between their 4th term is
(A) –1
(B) –8
(C) 7
(D) –9
2
Answer Key
1.
2.
3.
4.
5.
(C)
(A)
(D)
(B)
(Yes, A)
6.
7.
8.
9.
(B)
(B)
(A)
(C)
3
Hints and Solutions
1.
2.
3.
(C)
AP …. 17, 14, 11…–40
So, now we need the 6th term from last. So, by
reversing the AP, we get;
AP = –40, –37, –31…..14, 17
Now, given;
D=3
A1 = –40
n=6
Therefore a6 = a + (n – 1)d
= a6 = –40 + (6 – 1)3
= a6 = –40 + 5 * 3
= a6 = –40 + 15
= a6 = –25
So the 6th term is –25
(A)
So 100 will not be included in our case.
So writing the sequence we get,
= > 6, 8, 10, 12, 14 ……., 98
= > a = 6, d = 2, = 98, n = ?
Formula for calculating :
an = a + (n – 1)d
= > 98 = 6 + (n–1) 2
= > 98 = 6 = (n–1) 2
= > 92 = (n – 1) 2
= > 92/2 = (n – 1)
= > 46 = (n – 1)
= > n = 46 + 1 = 47
Hence there are 47 terms.
Now, we know that sum of n terms of an AP can be
given as:
S_n = n/2[a + an]
So substituting in the equation we get,
= > Sum of 47 terms = (47/2) (6 + 98)
= > Sum of 47 terms = (23.5) (104)
= > Sum of 47 terms = 2444
Hence the answer is 2444.
(D)
Series of numbers given is 21, 18, 15 ….
Here in this series of AP first term is 21.
We can calculate the common difference by
subtracting either 21 from 18 or 18 from 15.
d =18–21=15–18=–-3
So, the common difference is
–3
Therefore, a =
D = –3
Now it is given that the sum of the given AP series is
0. So, we have to equate the formula of the sum of n
number of terms in an AP equals to 0 and then by
solving we will get the value of n (number of terms).
Therefore, sum of n number of terms in an AP
= n/2[2a + (n – 1)d] = 0
On the other side then the equation becomes
 2a  (n  1)d  0  2a  (n  1)d  0
Now, we have to putting the value of a and d in the
above equation, we get
 2(21)  (n  1)(3)  0  2(21)  (n  1)(3)  0
Multiplying the terms, we get
 42  3n  3  0  42  3n  3  0
Now by simplifying the above equation we get
 3n  45  3n  45
n=15
Hence, 15 terms of the given AP series is added to get
the sum zero.
4.
(B)
5.
(Yes,A)
6.
(B)
7.
(B)
Let the first prize be a.
Then the prizes are
a, a – 10, a–20, ……… 10 terms
This is an A.P. with common difference d = –10
Sum of all the prizes = 800
n
 2a   n  1 d   800
2
10
 2  a  10  1 10   800
2
 2a  90  160
 2a  250
 a  125
8.
(A)
4
9.
(C)
Let the same common difference of two A.P. be d
First term of 1st AP = –1
Fourth term of 1st AP = (–1+3d)
First term of 2nd AP = –8
Fourth term of 2nd AP = (–8+3d)
Then (a4)1st AP – (a4)2nd AP = (–1+3d)–(–8+3d)
= –1+3d+8–3d=7
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