Question

Question 2 Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.

Class X1 - Maths -Sequences and Series Page 185

Answer 1

The numbers are:105, 110,115...995
Here, first term (a) = 105
and common difference (d) = 5
Let number of terms = n
Tₙ = a + (n-1)d
995 =105 + (n-1)5
890/5 = n-1
178 = n-1
n =179

now,
Sₙ =n/2{2a +(n-1)d}
=179/2{2×105 + (179-1)×5}
=179/2 × [ 210 + 178×5]
= 179/2 × [210+890]
= 179/2 × 1100
= 179 × 550
= 98450

Answer 2

Answer is in attachment.