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JEE Main 2026 Preparation: Question Papers, Solutions, Mock Tests & Strategy Unacademy » JEE Study Material » Mathematics » Sum to N Terms of Special Series

Sum to N Terms of Special Series

This lesson will let you find the sum to n terms of a special series. Sum to n terms of special series is very commonly seen in mathematics. This article will explain when to use how to get an accurate answer.

Table of Content
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When we explore the things we see in everyday life, it is not hard to spot a pattern or sequence. One puzzle that scientists puzzled over for many years was the sequence of sunflower petals. Research has found that these petals are part of Arithmetic Progression.

Previously, people would need a calculator to sum the integers from 1 to 100. Gauss came up with a process to calculate this and called it Arithmetic Progression. You could say that he is the father of this new way of calculating numbers.

In books VIII and IX of Euclid’s Elements, geometric progressions are analysed, giving their properties.

What is arithmetic progression :-

A sequence in which the differences between two consecutive numbers is constant. Such a sequence is said to be an arithmetic progression. 

An arithmetic progression is a set of numbers that follow simple, but strict rules. Every two successive terms in an A.P. are the same, and it is possible to see what the nth term will be by applying certain formulas.

In an arithmetic progression, the difference between each number will always be the same given the pattern.

The terms, except the first, are obtained by adding a set integer to the previous term in an arithmetic progression.

  • Sum of n terms in AP=n/2[2a + (n – 1)d]

  • Sum of natural numbers=n(n+1)/2

  • Sum of square of ‘n’ natural numbers=[n(n+1)(2n+1)]/6

  • Sum of Cube of ‘n’ natural numbers: [n(n+1)/2]2

What is geometric progression :-

A sequence in which the ratio between numbers remains constant. Such a sequence is known as geometric progression. 

A common pattern of progression is in geometric form, where each term has a fixed ratio. The progression is generally depicted as a, ar, ar2…. The adjectives α, π , and λ can be used to describe the progression.

Each successive term in a given geometric progression is obtained by multiplying the common ratio by the preceding term of that geometric progression.

Let us understand how GP sum works with an example: 

Simplifying complex concepts with GP sum is easy. For example, if someone saves $2 and increases the amount by $2 in each following week, then they will have $126 after 50 weeks. GP makes this process much Easier. 

If |r|<1, 

Sn=a(1−rn)/(1−r),

If |r|>1,Sn=a(rn−1)/(r−1)

The sum of the infinite geometric formula S∞=a/(1−r).

The formula for the nth  term of a geometric progression with the first term a and a common ratio of r =a1rn-1

Applications of Arithmetic Progression

  • Arithmetic Progression is used in straight-line depreciation calculation

  • Another way of how Arithmetic Progression is employed, is if someone needs to wait for a cab and they want to know when the next one will arrive, the A.P. will use their traffic algorithms to model on the next cab, more specifically when it will arrive.

  • Arithmetic Progression is used in pyramid-like patterns, with things changing constantly and many more applications.

  • With online banking, identity theft risk is high and it’s important to know basic math skills such as addition and subtraction. Checking accounts online is one example of where these skills are needed.

Applications of Geometric Progression:-

  • The height of the ball is reduced by half each time it bounces, until the ball stops. This is called a finite geometric sequence.

  • It is also used to calculate the compound interest. 

  • The Geometric progression formulae can also help to calculate the size of exponential population growth, such as bacteria in a Petri dish.

  • Assume one person was sick with the respiratory disease and did not cowl his mouth once 2 folks visited them when he was in bed. They leave, and therefore the illness affects them consecutive days. Imagine that every friend transmits the virus to 2 friends’ consecutive days by identical droplet distribution. If this pattern continues and every sufferer infects 2 a lot of folks, we will estimate the quantity of individuals infected.

Points to Remember:–

For AP

  • Common difference of an arithmetic progression d=a2−a1(Second term – First term)

  • Nth term of an arithmetic progression can be written as, an=a+(n−1)d.

  • Sum of nth  terms of an arithmetic progression Sn=n/2[2a+(n−1)d]

For GP

  • The common ratio of a geometric progression r=a2/a1

  • The formula for the nth  term of a geometric progression with the first term a and a common ratio of r =a1rn-1

  • The formula for the sum of the geometric progression is,

           If |r|<1,Sn=a(1−rn)1−r|r|<1,

           If |r|>1,Sn=a(rn−1)r−1|r|>1,

The sum of infinite geometric formula S∞=a1−r, where r<1.

faq

Frequently asked questions

Get answers to the most common queries related to the JEE Examination Preparation.

Find the 6th term of the G.P. 17, 204, 2448…

Ans: From the given a=17, r=(114/17)=12 We can use the formula r =a...Read full

Find the sum of the first 12 terms of the A.P.: 8,3,–2,…

Ans: Given AP: 8,3,−2,… Here a=8, common difference d=3−8=−5 an...Read full

The first term of an A.P. is 3, and the last term is 17. If the sum of all terms is 150, what is the 4th term?

Ans: From the given a=3,l=17,Sn=150,a...Read full

If the sum of the first 14 terms of an A.P. is 1050 and its first term is 10, find the 15th term.

Ans: Given: Sn=1050,n=14 and a=10 We know that the sum of n terms of an...Read full

Salary of Rohan, when his salary is ₹5,00,000 for the first year and expected to receive a yearly increment of 10. Now, find the Rohan salary at the start of the 5th year.

Ans: From the given information, we have a=500000,r=1.1 Rohan salary at...Read full

Ans: From the given a=17, r=(114/17)=12

We can use the formula r =a1rn-1

r=17(12)6-1=17(12)5= 4230144

We know the formula for the nth term of G.P.

Therefore, the 6th  term of the given G.P is 4230144.

Ans: Given AP: 8,3,−2,…

Here a=8, common difference d=3−8=−5 and n=12

We know,

Sum of n terms of an arithmetic progression Sn=n/2[2a+(n−1)d]

S=12/2[2(8)+(12−1)(−5)]

S=12/2[16+11(−5)]

S=6[16+11(−5)]

S=6[16−55]

S=6[−39]

S=−234

Therefore, the sum of the first 12 terms of the given A.P is −234.

Ans: From the given a=3,l=17,Sn=150,an= ?

Sn=n/2[a+l]

⇒150=n/2[3+17]

⇒150=n/2[20]

⇒150=10n

⇒n=150/10

⇒n=15

Now, a15=a+(n−1)d

⟹17=3+14d

⟹14d=14

⇒d=14/14=1

So, a4=a+3d

A4=3+3(1)

A4=3+3=6

Therefore, the 4th term is 6.

Ans: Given: Sn=1050,n=14 and a=10

We know that the sum of n terms of an arithmetic progression Sn=n2[2a+(n−1)d]

⇒1050=14/2[2(10)+(14−1)d]

⇒1050=7[20+13d]

⇒1050/7=20+13d

⇒150−20=13d

⇒13d=130

⇒d=130/13

⇒d=10

So, 15th  term of an arithmetic progression a15=10+(14−1)(10)

⇒a15=10+13(10)

⇒a15=10+130

⇒a15=140

Therefore, the 15th  term of an arithmetic progression is 140.

Ans: From the given information, we have a=500000,r=1.1

Rohan salary at the starting of 5th  year is needed. So, n=5

We know the formula for the nth  term of G.P.

We will get an=732050

Therefore, ₹732050 is the Rohan salary at the start of the 5th year.

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