# Mathematics - Class 10 / Grade 10

- Sum To n Terms of an Arithmetic Progression / Sum of the First n Terms of an Arithmetic Progression
### Sum To n Terms of an Arithmetic Progression / Sum of the First n Terms of an Arithmetic Progression

**Tags:**Arithmetic Sequences and Sums problem sums practice page for 10^{th}grade, Sum of N Terms of AP And Arithmetic Progression worksheet PDF for class X, Sum of the First n Terms of an Arithmetic Sequence examples for grade 10, How do you find the sum of Arithmetic Progression? How can you find the sum of n terms of the sequence without having to add all of the terms? For class tenth, How to find sum of n terms from nth term in an Arithmetic Progression?Sum To n Terms of an Arithmetic Progression

The sum of the first

**n**terms of an AP is given by**Note 1:**In the above formula, there are four quantities viz. S_{n}, a, n and d. If any three of these are known, the fourth can be determined. Sometimes two of these quantities are given, in such cases remaining two quantities are provided by some other relation.**Note 2:**If the sum S_{n}of n terms of a sequence is given, then nth term a_{n}of the sequence can be determined by the following formula.**a**_{n}= S_{n}– S_{n-1}**Example 1:**Find the sum of 10 terms of the A.P. 1, 4, 7, 10 ….**Solution:**Here values of a (first term) = 1, d (common difference) = 3, n (number of terms to be added) = 10We know that the sum of first

**n**terms of the Arithmetic Progress, whose first term = a and common difference = d is

**Example 2:**If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.**Solution:**Here S_{14}(sum of first 14 term) = 1050n (number of terms added) = 14

a (first term) =10

We know that the sum of first

**n**terms of the Arithmetic Progress, whose first term = a and common difference = d is**Example 3:**How many terms of the AP: 24, 21, 18 . . . must be taken so that their sum is 78?**Solution:**Here, a = 24, d = 21 – 24 = –3, S_{n}= 78. We need to find n.We know that the sum of first

**n**terms of the Arithmetic Progress, whose first term = a and common difference = d isThis implies the sum of the first 4 terms = the sum of the first 13 terms = 78.

Both values of ‘n’ are acceptable. So, the number of terms is either 4 or 13. This is because the sum of the terms from 5th to 13th will be zero as ‘a’ is positive and ‘d’ is negative that makes some terms positive and some negative which will cancel out each other.

**Example 4:**Find the sum of the first 100 positive integers.**Solution:**Let S = 1 + 2 + 3 + . . . + 100Using the formula for the sum of the first

**n**terms of an AP, we haveSo, the sum of the first 100 positive integers is 5050.

**Example 5:**Find the sum of first 20 terms of an A.P. whose nth term is given by a_{n}= 2 + 3n**Solution:**We have, a_{n}= 2 + 3nTherefore,

a

_{1}= 2 + 3x1 = 5a

_{2}= 2 + 3x2 = 8a

_{3}= 2 + 3x3 =11 and so ond = a

_{2}-a_{1}= 8-5 =3To find S

_{20}, we have a = 5, d = 3, n = 20.So, the sum of first 20 terms is 670.

**Example 6:**A manufacturer of electric car produced 500 cars in the second year and 700 cars in the sixth year. Assuming that the production increases uniformly by a fixed number every year, find the production in the 1st year and the total production in first 6 years**Solution:**Here, a_{2}= 500 and a_{6}= 700Or we write a + d = 500 and a + 5d = 700

Therefore, production of electric car in the first year is 50.

Thus, the total production of electric car in first 6 years is 1050

**Example 7:**Find the sum of all natural numbers between 500 and 1000 which are exactly divisible by 3.**Solution:**Clearly, the numbers between 250 and 1000 which are divisible by 3 are 501, 504, 507 ….. 999. This is an A.P. with first term a = 501, common difference = 3 and last term = 999. Let there be n terms in this A.P. Then,**Download to practice offline.**