Question

Find the sum of the integers between 100 and 200 that are (i) divisible by 9 (ii) not divisible by 9.

Answer 1

Hint:Here, for the first part write all terms between 100 to 200 divisible by 9, these terms form an A.P. Using formula find the number of terms and sum of all terms. For the second part, find the sum of all terms between 100 to 200, and from the result subtract the result obtained in part (i).

Complete step by step answer:
(i) Numbers between 100 and 200 divisible by 9 are 108, 117, 126, …, 198.We have a sequence: 108, 117, 126, …, 198.In which, first term is 108, common difference = 117 – 108 = 9, last term = 198
For number of terms,
$[L = a + (n – 1) × d]\\
\Rightarrow 198 = 108 + (n − 1) × 9 \\
\Rightarrow 90 = (n − 1) × 9 \\
\Rightarrow (n − 1) = 10 \\
\Rightarrow n = 11$
Sum of series = $\dfrac{n}{2}$ ​[first term + last term]
Sum of series = $\dfrac{{11}}{2} ​[108 + 198] = 11 × (306) = 3366$

(ii) All terms between 100 to 200 are 101, 102, 103, …, 199
This is an A.P. with first term 101, common difference = 1, last term = 199
For number of terms,
$[L = a + (n – 1) × d]\\
\Rightarrow 199 = 101 + (n − 1) × 1 \\
\Rightarrow 98 = (n − 1) \\
\Rightarrow n = 99$
Sum of series = $\dfrac{n}{2}$ ​[first term + last term]
Sum of series = $\dfrac{{99}}{2} ​[101 + 199] = 99 × (150) = 14850$
Sum of integers between 100 to 200 not divisible by 9 is sum of all terms between 100 to 200 – sum of all terms divisible by 9
$\Rightarrow 14850 – 3366 = 11484$

Note: In these types of questions, find the terms according to the given conditions and check whether the terms form A.P. or G.P, apply formula to find the sum of terms.Arithmetic Progression (AP) is a sequence of numbers in order in which the difference of any two consecutive numbers is a constant value.