This set of Aptitude Questions and Answers (MCQs) focuses on “Progression”.
1. If the sum of n terms of an AP is (3n2 + n) and Tm = 166, then what is the value of m?
a) 27
b) 28
c) 29
d) 26
View Answer
Explanation: Given that Sn = (3n2 + n) and Tm = 166
We know that, Tm = Sm – Sm-1
166 = 3m2 + m – 3(m-1)2 – (m-1)
166 = 6m – 2
Hence, m = 28
2. What will be the maximum sum of 72, 69, 66, ……?
a) 900
b) 897
c) 882
d) 903
View Answer
Explanation: The sequence is in AP with a = 72, d = -3.
For the sum to be maximum, the sequence should never take negative values. Hence, the last term of sequence is 0 i.e., Tn = 0
Tn = a + (n-1)d
0 = 72 + (n – 1)(-3)
n = 25
Sum of an AP, Sn = \((\frac{first \, term+last \, term}{2})\)*number of terms
Sn = \(\frac{72+0}{2}\) * 25 = 900
3. In a GP, if the (m+n)th term is x and (m-n)th term is y, then find its mth term.
a) √(xy)
b) xy
c) x/y
d) √x/√y
View Answer
Explanation: Tm+n = arm+n-1 = x …… (i)
Tm-n = arm-n-1 = y …… (ii)
On multiplying (i) and (ii), we get
a2r2m-2 = xy
Tm = arm-1 = √(xy)
4. How many numbers are there in between 55 and 3815, which are multiples of 5 and divisible by 3?
a) 250
b) 251
c) 249
d) 238
View Answer
Explanation: For a number to be a multiple of 5 and divisible by 3, it should be divisible by 15.
Between 55 and 3815, the first and last number divisible by 15 are 60 and 3810 respectively.
i.e., a=5, d=15, Tn=3810
Tn = a + (n-1)d
3810 = 60 + (n-1)*15
Therefore, n=251
5. The sum of 3 numbers in AP is 48. The sum of their squares is 800. Find the largest among the three numbers.
a) 12
b) 16
c) 20
d) 24
View Answer
Explanation: Let the three numbers in AP be a-d, a, a+d
Given, sum of those numbers is 48.
i.e., (a-d) + a + (a+d) = 48
Hence, a=16
Also, given that sum of their squares is 800.
i.e., (a-d)2 + a2 + (a+d)2 = 800
(16-d)2 + 162 + (16+d)2 = 800
Hence, d=4
Therefore, the largest number (a+d) is 20.
6. Find the sum of the three numbers in G.P whose product is 512 and the sum of their products taken in pairs is 224.
a) 28
b) 24
c) 20
d) 16
View Answer
Explanation: Let the three numbers in G.P be a/r, a, ar.
Given that product of these numbers is 512.
i.e., \(\frac{a}{r}\)*a*ar=512
a = 8
Also, given that sum of their products taken in pairs is 224.
i.e., \(\frac{a}{r}\)*a+a*ar+\(\frac{a}{r}\)*ar=224
Hence, r = 2
The three numbers are 4, 8, 16
Therefore, the sum of three numbers in GP is 28.
7. The third term of a G.P is 5. Find the product of first five terms.
a) 53
b) 5
c) 55
d) 0
View Answer
Explanation: Let the five numbers in G.P be a/r2, a/r, a, ar, ar2
Given that, third term i.e., a=5
Then the product of five terms is,
\(\frac{a}{r^2} * \frac{a}{r}\)*a*ar*ar22=a5
Therefore, the product of first five terms is 55.
8. The sum of n terms of an AP is (3n2+5n), then which term of AP is 224?
a) 34th term
b) 32th term
c) 39th term
d) 37th term
View Answer
Explanation: We know that, nth term of AP is given by an = Sn – Sn-1
an = 3n2 + 5n – 3(n-1)2 – 5(n-1)
an = 6n + 2
Given that an=224
Therefore, n = 37
9. If m, 2m-1 and 2m+1 are three consecutive terms of AP, then what is the value of m?
a) -3
b) 3
c) 6
d) 4
View Answer
Explanation: For a sequence to be in AP, difference between any two consecutive terms must be same.
Hence, (2m – 1) – m = (2m + 1) – (2m – 1)
Therefore, m = 3
10. The sum of second and seventh term of an AP is 6. Find the sum of the first 8 terms.
a) 24
b) 34
c) 21
d) 29
View Answer
Explanation: Given that, T2 + T7 = 6
a + d + a +6d = 6
2a + 7d = 6 …… (i)
Sum of first n terms is Sn = \(\frac{n}{2}\)(2a+(n-1)d)
S8 = \(\frac{8}{2}\)(2a+7d) = \(\frac{8}{2*6}\) …… (from (i))
Therefore, S8 = 24
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