Short Answer TypeFind the sum of all natural numbers lying between 100 and 1000, which are mutliples of 5.
All the natural numbers between 100 and 1000 which are multiples of 5 are:
105, 110, 115, ......, 995.
Let 
denote their sum, i.e., 
Here, a = 105, d = 110 - 105 = 5, 
i.e., a + (n - 1)d = 995 or 105 + (n - 1)5 = 995
or 5n + 100 = 995 or 5n = 995 - 100 = 895
or 
∴ 
Find the sum of all the two digit numbers, which when divided by 4, yields 1 as remainder.
The sum of first 4 terms of an A.P. is 56. The sum of last 4 terms is 112. If its first term is 11, then find the number of terms.
The sums of first n terms of two A.P.’s are in the ratio of 14 – 4n : 3n + 5. Find the ratio of their 8th terms.
The ratio of sums of m and n terms of an A.P. is m2 : n2. Show that the ratio of with and nth term is 2m – 1 :2n – 1.
Long Answer TypeIf S1, S2, S3 ........Sp are the sums of n terms of p A.P.’s whose first terms are 1, 2,3, .....p and common differences are 1, 3, 5, .....(2p – 1) respectively. Show that
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