In AP (Arithmetic Progression), the difference between two consecutive terms remains constant.

The nth term 'T(n)' of an AP is given by:

T(n) = a + (n-1)d; where a is the initial term and d is the common difference.

â€‹

The sum of n terms S(n) of an AP is given by:

S(n) = (n/2){2a + (n-1)d}; or, (n/2){a + l}

where l is the last term of AP, as for last term, it is equal to a+(n-1)d

â€‹

Arithmetic Progression

Objectives

This Chapter introduces you to the concepts of Arithmetic Progression.

Qu. 01: How many three-digit numbers are divisible by 7?

Sol: The first three-digit numbers divisible by 7 is 105. Obviously the last three-digit number divisible by 7 is 994, hence the required AP is 105, 112, 119, …, 994 with the common difference d = 7.

Let the last three-digit number 994 be nth term, i.e. T(n) of this AP. Then, T(n) = a + (n-1)d

994 = 105 + (n-1)7, 7n – 7 = 889; 7n = 896; n = 128

Thus there are 128 three-digit numbers divisible by 7.

Qu. 02: Determine the AP whose third term is 16 and seventh term exceeds the fifth term by 12?

Sol: Given, T(3) = 16; and T(7) – T(5) = 12. Thus, if a is the initial term and d is the common difference, then:

16 = a + (3-1)d or a + 2d = 16; and

12 = {a + (7-1)d} – {a + (5-1)d}; 2d = 12, d = 6. Hence, a = 16 – 2d = 16 – 12 = 4.

The AP is: 4, 10, 16, 22, 28, 34, 40, 46, ..

Qu. o3: An AP consists of 50 terms of which third term is 12 and the last term is 106. Find the 29th term.

Sol: Let a be the initial term and d the common difference. Given, T(3) = a + (3-1)d = 12;

i.e. a+2d = 12

And, T(50) = a + (50-1)d = 106; i.e. a + 49d = 106

On subtraction of first equation from the second one, we have, 47d = 94, d = 2

Thus, a = 12 – 2d = 12 – 4 = 8.

Thus, T(29) = a + (29-1)d = 8 + (28x2) = 8+56 = 64.

Qu. o4: If the third and ninth terms of an AP are 4 and -8 respectively, which term of this AP is zero?

Sol: Given, T(3) = 16; and T(7) – T(5) = 12.

Thus, if a is the initial term and d is the common difference, then:

16 = a + (3-1)d or a + 2d = 16; and

12 = {a + (7-1)d} – {a + (5-1)d}; 2d = 12, d = 6.

Hence, a = 16 – 2d = 16 – 12 = 4.

The AP is: 4, 10, 16, 22, 28, 34, 40, 46, ..

Qu. 05: Justify if it is true to say that the following are nth terms of an AP. (i) 2n – 3 (ii) 3n^2 + 5 (iii) 1+n+n^2

Sol: A linear explanation of nth term, like pn+q where p and q are rational numbers can always be put in the form: (q+p) + (n-1)p; which is nth term of an AP with initial term being (q+p) and common difference p.

On the other hand, a quadratic explanation of nth term, like pn^2+rn + q where p, r and q are rational numbers can never be put in the form: a + (n-1)d, as there shall be an exponential increment in every consecutive term due to n^2., hence not an AP.

Hence, for given question, (i) is an AP, while (ii) and (iii) are not AP.

Otherwise also,

For (i), T(1) = -1, T(2) = 1, T(3) = 3, T(4) = 5; this is an AP with common difference 2 and initial term equal to -1.

For (ii), i.e. 3n^2 + 5, we have, T(1) = 8, T(2) = 17, T(3) = 32. Obviously, T(3) – T(2) is not equal to T(2) – T(1); this is NOT an AP.

For (iii), i.e. 1+n+n^2, we have, T(1) = 3, T(2) = 7, T(3) = 13. Obviously, T(3) – T(2) is not equal to T(2) – T(1); this is NOT an AP.

â€‹

This explanation is available on my blog too.

Qu. 06: Two APs have the same common difference. The difference between their 100th terms is 100. What is the difference between their 1000th terms?

Sol: Given, T2(100) – T1(100) = 100; i.e. {a + (100-1)d} – {b + (100-1)d} = 100, where a and b are initial terms of two APs and d is the same common difference. Thus, a – b = 100

Hence, T2(1000) – T1(1000) = {a + (1000-1)d} – {b + (1000-1)d} = a-b = 100. The difference between their 1000th terms is 100.

Remember, for two APs, with initial terms being a and b respectively, and common difference being same, say d, the difference between their nth terms remains same, and is equal to a – b, as, {a + (n – 1)d} – {b + (n – 1)d} = a-b, the constant.

â€‹

This explanation is available on my blog too.