# ARITHMETIC PROGRESSION QUESTIONS FOR CA FOUNDATION

Problem 1 :

If the general term of a sequence is (n - 1)/(n + 1), find the 3rd term of the sequence.

A)  -2     B)  1/3     C)  1/2    D)  3

Solution :

tn = (n - 1) / (n + 1)

When n = 3

t3 = (3 - 1) / (3 + 1)

t3 = 2/4

t3 = 1/2

So, option C is correct.

Problem 2 :

Which of the following is not an arithmetic sequence?

A)  11, 2, -8, -19, ..........        B)  4, 7, 10, 13, ............

C)  57, 51, 45, 39, ........     D)  -3, -5, -7, -9, .......

Solution :

Figuring common difference from option a :

11, 2, -8, -19, ..........

2 - 11 ==> -9

-8 - 2 ==> -10

Since the common difference is not same, it is not arithmetic progression.

Problem 3 :

If the common difference of an arithmetic sequence is 4 and T(6) = 15, find the first term of the sequence.

A)  -9     B)  -5     C)  11/5     D)  35

Solution :

Common difference (d) = 4

t6 = 15

a + 5d = 15

Applying the value of d, we get

a + 5(4) = 15

a + 20 = 15

a = 15 - 20

a = -5

So, option B is correct.

Problem 4 :

If the general term of an arithmetic sequence is T(n) = 7 - 3n, find the sum of the first 7 terms of the sequence.

A)  -56    B)  -35     C)  4    D)  210

Solution :

T(n) = 7 - 3n

To find a, let us apply n = 1

T(1) = 7 - 3 ==> 4 = a

To find d, we apply n = 2

T(2) = 7 - 6 ==> 1

Common difference (d) = 1 - 4 ==> -3

Sum of first 7 terms = (n/2) (2a + (n - 1) d)

= (7/2) [2(4) + (7 - 1)(-3)]

= (7/2) [8 + 6(-3)]

= (7/2) [8 - 18]

= 7(-5)

= -35

So, option B is correct.

Problem 5 :

Find the sum of the integers between 301 and 400 inclusive that are multiples of 4.

A)  8400    B)  8800     C)  20200    D)  49900

Solution :

304 + 308 +........... + 400

Number of terms in the above series.

Problem 6 :

Find the sum of the arithmetic series.

Solution :

Problem 7 :

The sum of all natural numbers between 100 and 1000 which are multiple of 5 is.

A)  99550     B)  96450      C)  97450      D)  95450

Solution :

Problem 8 :

The sum of an AP whose first term is -4 and last term is 146 is 7171. Find the value of n.

A)  99    B)  100      C)  101     D)  102

Solution :

a = -4, l = 146 and sn = 7171

Problem 9 :

Divide 30 into five parts in A.P such that the first and last parts are in the ratio 2 : 3

Solution :

First part = 24/5, last part = 36/5

First part : last part = 24/5 : 36/5

= 24 : 36

= 2 : 3

Problem 10 :

Insert 4 A.M' between 3 and 18.

A)  12, 15, 9, 6        B)  9, 6, 12, 15       C)  6, 9, 12, 15

D)  15, 12, 9, 6

Solution :

3, a+d, a+2d, a+3d, a + 4d, 18

a+5d = 18

Applying the value of a, we get

3+5d = 18

5d = 15

d = 3

 2nd term6 3rd term9 4th term12 5th term15

So, option B is correct.

Problem 11 :

(x + 1), 3x and (4x + 2) are in A.P. Find the value of x.

A)  2    B)  3    C)  4   D)  5

Solution :

3x - (x + 1) = (4x + 2) - 3x

3x - x - 1 = 4x - 3x + 2

2x - 1 = x + 2

2x - x = 2 + 1

x = 3

So, option B is correct.

Problem 12 :

Find the sum of the series

2 + 7 + 12+ ...........297

A)  8970    B)  7630    C)  8870     D) 9875

Solution :

a = 2, d = 7 - 2 ==> 5

Number of terms :

n = (l - a)/d + 1

n = (297 - 2)/5 + 1

n = 295/5 + 1

n = 60

S60 = (60/2)[2 + 297]

= 30(299)

= 8970

Problem 13 :

Find the number which should be added to the sum of any number of terms of the A.P

3, 5, 7, 9, 11, ......

resulting in a perfect square,

A)  -1    B)  0    C)  1      D)  none

Solution :

3 + 5 ==> 8 (not a perfect square)

3 + 5 + 7 ==> 15 (not a perfect square)

3 + 5 + 7 + 9 ==> 24

To make the sum as perfect square, we have to add 1. So, option C is correct.

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