Here we will learn how to solve different types of problems on sum of n terms of Arithmetic Progression.

**1.** Find the sum of the first 35 terms of an Arithmetic Progression whose third term is 7 and seventh term is two more than thrice of its third term.

**Solution:**

Let us assume that ‘a’ be the first term and ‘d’ be the common difference of the given Arithmetic Progression.

According to the problem,

3rd term of an Arithmetic Progression is 7

i.e., 3th term = 7

⇒ a + (3 - 1)d = 7

⇒ a + 2d = 7 ................... (i)

and seventh term is two more than thrice of its third term.

i.e., 7th term = 3 × 3rd term + 2

⇒ a + (7 - 1)d = 3 × [a + (3 - 1)d] + 2

⇒ a + 6d = 3 × [a + 2d] + 2

Substitute the value of a + 2d = 7 we get,

⇒ a + 6d = 3 × 7 + 2

⇒ a + 6d = 21 + 2

⇒ a + 6d = 23 ................... (ii)

Now, subtract the equation (i) from (ii) we get,

4d = 16

⇒ d = \(\frac{16}{4}\)

⇒ d = 4

Substitute the value of d = 4 in the equation (i) we get,

⇒ a + 2 × 4 = 7

⇒ a + 8 = 7

⇒ a = 7 - 8

⇒ a = -1

Therefore, the first term of the Arithmetic Progression is -1 and common difference of the Arithmetic Progression is 4.

Now, sum of the first 35 terms of an Arithmetic Progression S\(_{35}\) = \(\frac{35}{2}\)[2 × (-1) + (35 - 1) × 4], [Using the Sum of the First n Terms of an Arithmetic Progression S\(_{n}\) = \(\frac{n}{2}\)[2a + (n - 1)d]

= \(\frac{35}{2}\)[-2 + 34 × 4]

= \(\frac{35}{2}\)[-2 + 136]

= \(\frac{35}{2}\)[134]

= 35 × 67

= 2345.

**2.** If the 5th term and 12th term of an
Arithmetic Progression are 30 and 65 respectively, find the sum of its 26
terms.

**Solution:**

Let us assume that ‘a’ be the first term and ‘d’ be the common difference of the given Arithmetic Progression.

According to the problem,

5th term of an Arithmetic Progression is 30

i.e., 5th term = 30

⇒ a + (5 - 1)d = 30

⇒ a + 4d = 30 ................... (i)

and 12th term of an Arithmetic Progression is 65

i.e., 12th term = 65

⇒ a + (12 - 1)d = 65

⇒ a + 11d = 65 .................... (ii)

Now, subtract the equation (i) from (ii) we get,

7d = 35

⇒ d = \(\frac{35}{7}\)

⇒ d = 5

Substitute the value of d = 5 in the equation (i) we get,

a + 4 × 5 = 30

⇒ a + 20 = 30

⇒ a = 30 - 20

⇒ a = 10

Therefore, the first term of the Arithmetic Progression is 10 and common difference of the Arithmetic Progression is 5.

Now, sum of the first 26 terms of an Arithmetic Progression S\(_{26}\) = \(\frac{26}{2}\)[2 × 10 + (26 - 1) × 5], [Using the Sum of the First n Terms of an Arithmetic Progression S\(_{n}\) = \(\frac{n}{2}\)[2a + (n - 1)d]

= 13[20 + 25 × 5]

= 13[20 + 125]

= 13[145]

= 1885

**●** **Arithmetic Progression**

**Definition of Arithmetic Progression****General Form of an Arithmetic Progress****Arithmetic Mean****Sum of the First n Terms of an Arithmetic Progression****Sum of the Cubes of First n Natural Numbers****Sum of First n Natural Numbers****Sum of the Squares of First n Natural Numbers****Properties of Arithmetic Progression****Selection of Terms in an Arithmetic Progression****Arithmetic Progression Formulae****Problems on Arithmetic Progression****Problems on Sum of 'n' Terms of Arithmetic Progression**

**11 and 12 Grade Math**__From Problems on Sum of 'n' Terms of Arithmetic Progression____ to HOME PAGE__

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