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Arithmetic Progressions

Question
CBSEENMA10009823
Wired Faculty App

Find the sum of all three digit natural numbers, which are multiples of 7.

Solution

First three-digit number that is divisible by 7 = 105

Next number = 105 + 7 = 112

Therefore the series is  105, 112, 119,........

The maximun possible three digit number is 999.

When we divide by 7, the remainder will be 5.

Clearly, 999 - 5 = 994 is the maximum possible three-digit number divisible by 7.

The series is as follows:

105, 112, 119,.........994.

 Here  a = 105,  and  d = 7

Let 994 be the nth term of this A.P.

      an = a + ( n-1 ) d

⇒ 994 = 105 + ( n-1 ) x 7

⇒  ( n-1 ) x 7 = 889

⇒  ( n-1 ) = 127

⇒   n = 128

So, there are 128 terms in the A.P.

∴ Sn = n2First term + last term                     = 1282 a1 + a128           = 64 ( 105 + 994 )           = 64 x 1099           = 70336

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, . . .

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