50 + 46 + 42 + 38 +... to n terms = 336.

Find the sum of the following arithmetic progressions: (i) 50 ,\ 46 ,\ 42 ,\ ... to 10 terms (ii) 1,\ 3,\ 5,\ 7,\ ... to 12 terms (iii) 3,\ 9//2,\ 6,\ 15//2,\ ... to 25 terms

2+ 3.2 + 4.2^(2) + …... upto n terms =

Consider the arithmetic sequence -46,-42,-38,….., How many terms added from first to get the sum zero?

42 40 38 35 33 31 28 ?

Find the sum of the following arithmetic progressions: 50,46,42, to 10 terms (ii) 1,3,5,7, to 12 terms (iii) 3,9/2,6,15/2,dots to 25 terms

Find the term of an AP whose term is 38 and the term is 38 and the term is 73.

Let S_n, be the sum of n terms of 40, 38, 36, 34 ......., then S_n, can be: (A) 310 (B) 180 (C) 330 (D) -42

In an A.P., 1^(st) terms is 1 and the last term is 20. The sum of all terms is 399 then n = …………. A)42 B)38 C)21 D)19

In an A.P, 1st term is 1 and the last term is 20 .The sum of all terms is 399 , then n=…. a) 42 b) 38 c) 21 d) 19

A compound contains 38. 8 % C, 16 % H, 42 5 % N. The formula of compound will be: