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
11 and 12 Grade Math
From Problems on Sum of 'n' Terms of Arithmetic Progression to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.

New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.