SBI Sample Arithmetic Progression Problems

Dear Reader,
Below are four problems based on A.P series

Question 1

The n-th term of the series 6,10,14,... is:
a) 6n-1 b) 4n-2 c) 2(2n+1) d) 6n+1

Answer : c) 2(2n+1)

Solution :

The given series 6,10,14,... is in the form of arithmetic series
Here, the first term a = 6, common difference d = 4
The n-th term = tn = a + (n-1)d
tn = 6 + (n-1)4 = 6 + 4n - 4 = 4n + 2 = 2(2n+1)

Hence the answer is 2(2n+1)

Question 2

The sum of the first 100 - terms of the series 1/2, 3/2, 5/2, 7/2,... is:
a) 1000 b) 2000 c) 5000 d) 10000

Answer : c) 5000

Solution :

The given series is in the form of arithmetic progression.
Here, a = first term = 1/2, common difference = d = 1

we have to find the sum of the n terms.
We know that, " the sum of the first n-terms of the A.P series = n/2 (2a + (n-1)d).
Now, the sum of 100 terms = 100/2 [(2x1/2) + (99)1]
= 50 [ 1 + 99] = 50 x 100
= 5000.
Hence the answer is 5000.

Question 3

Find the total number of terms in 10,34,58,...,466
a) 30 b) 23 c) 25 d) 20

Answer : d) 20

Solution:

The given series 10,34,58,...,466 is in the form of arithmetic series.
Here a = first term = 10, d = common difference = 24 and l = last term = 466.
We know that the number of terms in A.P = n = (l-a)/d + 1
n = (466 - 10)/ 24 + 1
= 456 / 24 + 1
= 19 + 1
= 20.

Hence, the series has 20 terms.

Question 4

Suppose the series 28, 25, ...., -29 has 20 terms then the sum of all 20 terms is equal to:
a) -5 b) 10 c) -10 d) 5

Answer : c) -10

Solution:

Given series is 28, 25, ...., -29 and it has 20 terms with a = 28 and d = -3, l = -29 and n = 20

The sum of all n-terms in arithmetic sequence with last term(l) = S(n) = n(a+l)/2

Then, s(20) = 20(28 + (-29)) / 2
= 20(-1) / 2
= -10
Hence the required sum is -10.


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