+91-8076753736[email protected]
logo
logo
-->

Arithmetic Progressions

Question
CBSEENMA10007874
Wired Faculty App

Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Solution

Here n = 51, a2 = 14, a3 = 18
d = a3 – a2 = 18 – 14 = 4 a2 = a + d
⇒ 14 = a + 4
⇒ a = 14 – 4 = 10

We know that,
     straight S subscript straight n equals straight n over 2 left square bracket 2 straight a plus left parenthesis straight n minus 1 right parenthesis straight d therefore space straight S subscript 51 equals 51 over 2 left square bracket 2 cross times 10 plus left parenthesis 51 minus 1 right parenthesis cross times 4 right square bracket equals 51 over 2 left square bracket 20 plus 50 cross times 4 right square bracket equals 51 over 2 cross times left parenthesis 20 plus 200 right parenthesis equals 51 over 2 cross times 220 equals 51 cross times 110 equals 5610.

Some More Questions From Arithmetic Progressions Chapter

In which of the following situational, does the list of numbers involved make an arithmetic progression, and why?

The cost of digging a well for the first metre is Rs. 150 and rises by Rs. 50 for each succeeding metre.

Write first four terms of the AP, when the first term a and the common difference d are given as follows:
a – 10, d = 10

Write first four terms of the AP, when the first term a and the common difference d are given as follows:
a = –2, d = 0

Write first four terms of the AP, when the first term a and the common difference d are given as follows:
a = 4, d = – 3

Write first four terms of the AP, when the first term a and the common difference d are given as follows:

a = – 1.25, d = – 0.25

For the following APs, write the first term and the common difference
3, 1, – 1, – 3, . . .

For the following APs, write the first term and the common difference
– 5, – 1, 3, 7, . . .

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation