Progressions - Model Problems - 7

1. The third term of a geometric progression is 4. The product of the first five terms is:

a. 4³
b. 4⁴
c. 4⁵
d. none of these

(JEE 1982)

Answer ©

Select the five terms as a/r², a/r,a, ar,ar².

Product of the five terms is a/r² * a/r * a *ar*ar² = a⁵

As the third term is 4, a = 4
Hence product is 4⁵

2. The sum of integers from 1 to 100 that are divisible by 2 or 5 is ---------------------.
(JEE 1984)

Answer: 3050

Integers divisible by 2 are 2,4,6,…,100 ….(50 numbers)
Integers divisible by 5 are 5, 10, 15,…100 (20 numbers)

Integers divisible by 10 will be in integers divisible by 2. We can remove them from integers divisible by 5 to find out the sum required

So integers for which sum is to be taken 2,4,6…,100 and
5,15,25,…,95

First progression sum 2(sum of 1,2,3,…,50)
= 2*50*51/2 = 2550

Second progression is an A.P. with d = 10 and a = 5

Hence sum = n/2(a+l) = 10/2(100) = 500

Total sum = 2550 +500 = 3050