Question

Pls any one and these questions I will mark u as brainlest

Answer 1

Answer:

option D is the answer for the given problem

Answer 2

Ans for 1.6:- A.  3

Step-by-step explanation:

since they are the consecutive terms of AP,

4k - 6 - (k + 2) = 3k - 2 - (4k - 6)

or, 4k -  6 - k - 2 = 3k - 2 - 4k + 6

or, 4k - k - 3k + 4k = 6 + 2 - 2 + 6

or, 4k = 12

therefore, k = 3

Ans for 1.7:- D.  3775

given,

first term, a = 51

last term, l = 100

common difference,d = 52 - 51 = 1

sum ,Sn = ?

total no, Tn =?

now,

l = a + (n-1)d

or, 100 = 51 + (n - 1) * 1

or, 100 = 51 + n - 1

or, 100 - 50 = n

therefore, n = 50

then,

$S_{50}$ = $\frac{n}{2}$[2a + (n - 1) d]

     = $\frac{50}{2}$[2 * 51 + (50- 1) *1]

     = 25[102 + 49]

     = 25 * 151

     = 3775

Ans for 1.8:-  C.  5910