## 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.