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If the sum of first p terms of an AP is (ap² + bp), find its common difference?

  • Last Updated : 01 Sep, 2021

Arithmetic probably has the longest history during the time. It is a method of calculation that is been in use from ancient times for normal calculations like measurements, labeling, and all sorts of day-to-day calculations to obtain definite values. The term got originated from the Greek word “arithmos” which simply means numbers.

Arithmetic is the elementary branch of mathematics that specifically deals with the study of numbers and properties of traditional operations like addition, subtraction, multiplication, and division.

Besides the traditional operations of addition, subtraction, multiplication, and division arithmetic also include advanced computing of percentage, logarithm, exponentiation and square roots, etc. Arithmetic is a branch of mathematics concerned with numerals and their traditional operations.

What is an Arithmetic Progression?

Arithmetic Progression, (AP) is a sequence of numbers in which the difference between any two consecutive numbers is a constant value. For example, the series of numbers: 1, 2, 3, 4, 5, 6,… are in Arithmetic Progression, which has a common difference (d) between two successive terms (say 1 and 2) is equal to 1 (2 – 1).

A common difference can be seen between two successive terms, even for odd numbers and even numbers. In AP, three main terms are Common difference (d), nth Term (an), Sum of the first n terms (Sn). All three terms represent the properties of AP. Let’s take a look at what common difference is in detail: 

In other words, arithmetic progression can be defined as “A mathematical sequence in which the difference between two consecutive terms is always a constant“.

We come across different words like sequence, series, and progression in AP, now let us see what does each word define:

Sequence is a finite or infinite list of numbers that follows a certain pattern. For example 0, 1, 2, 3, 4, 5… is the sequence, which is an infinite sequence of whole numbers.

Series is the sum of the elements in which the sequence is corresponding. For example 1 + 2 + 3 + 4 + 5…. is the series of natural numbers. Each number in a sequence or a series is called a term. Here 1 is a term, 2  is a term, 3 is a term ……. and so on.

Progression is a sequence in which the general term can be expressed using a mathematical formula or the Sequence which uses a mathematical formula that can be defined as the progression.

General form of arithmetic progression is  

a, a + d, a + 2d, ………a + (n – 1) d

Here are some examples of AP:

  • 6, 13, 20, 27, 34, 41, . . . .
  • 91, 81, 71, 61, 51, 41,. . . .
  • π, 2π, 3π, 4π, 5π,6π ,…
  • -√3, −2√3, −3√3, −4√3, −5√3, – 6√3,…..

Common Difference of an A.P.

The common difference is denoted by d in arithmetic progression. It’s the difference between the next term and the one before it. For arithmetic progression, it is always constant or the same. In a word, if the common difference is constant in a certain sequence, we can say that this is A.P. If the sequence is a1, a2, a3, a4, and so on.

In other words, the common difference in the arithmetic progression is denoted by d. The difference between the successive term and its preceding term. It is always constant or the same for arithmetic progression. In other words, we can say that, in a given sequence if the common difference is constant or the same then we can say that the given sequence is in Arithmetic Progression (AP).

The formula to find common difference is d = (an + 1 – an) or d = (an – an-1).

If the common difference is positive, then AP increases. For Example 4, 8, 12, 16….. in this series, AP increases

If the common difference is negative then AP decreases. For Example -4, -6, -8……., here AP decreases.

If the common difference is zero then AP will be constant. For Example 1, 1, 1, 1, 1………, here AP is constant.

The sequence of Arithmetic Progression will be like a1 , a2 , a3 , a4  ,…

common difference  (d) = a2 – a1 = d  

                                            a3 – a2 = d

                                            a4 – a3 = d and so on.

If the sum of first p terms of an AP is (ap² + bp), find its common difference?

Answer:

Sum of pth term of an ap is ap2 + bp 

i.e., sp = ap2 + bp

Let’s assume the value of p is 1 for s1

So, sp = ap2 + bp

s1 = a (1)2 + b(1)

s1  =  a + b ⇒  a1   1st term

Therefore s1 = a + b

same for sp = ap2 + bp   {p = 2}

              s2 = a(2)2 + b(2)

                  = 4a + 2b

now for a2 = s2 – s1

                  =  4a + 2b – (a+b)

                  = 4a + 2b – a – b

              a2 = 3a + b ⇒ 2nd term

now the common difference 

d = a2 – a1

   = 3a + b  – ( a+b)

   = 3a + b – a – b

d = 2a

Therefore the common difference is d = 2a

Similar Questions

Question 1: Calculate the common difference of the AP 1/b, 3–b/3b, 3–2b/3b…

Solution:  

Here the AP is 1/b, 3-b/3b, 3-2b/3b

so as we now take a term as

                                                             a1 = 1/b

                                                             a2 =  3-b/3b

                                                             a3 = 3-2b/3b

So the formula to find the common difference is

                                                             d = a2 – a1

                                                                = 3-b/3b – 1/b

                                                                = (3 – b – 3)/3b

                                                                = -b/3b

therefore,                                             d  = -1/3

So the common difference of 1/b, 3 – b/3b, 3 – 2b/3b is – 1/3

Question 2: Find the AP if the first term is 25 and the common difference is 4.

Solution:

As we know,

a, a + d, a + 2d, a + 3d, a + 4d, …

Here, a = 25 and d = 4

                                = 25, (25 + 4), (25 + 2 × 4), (25 + 3 × 4), (25 + 4 × 4),

                                = 25, 29, (25 + 8), (25 + 12), (25 + 16), …

                                = 25, 29, 33, 37, 41, …and so on.

So the AP is 25, 29, 33, 37, 41…..

Question 3: Find the 30th term for the given AP:  7, 9, 11, 13, 15 …

Solution:  

Given 7, 9, 11, 13, 15……

here,          

                      a = 7, d = 5 – 3 = 2, n = 30

                      an = a + (n − 1)d

                     a30 = 7 + (30 − 1)2

                    a30  = 7 + 58

                    a30 = 65

Here 30th term is a30 = 65


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