Math, asked by thamimbinmuneer, 10 months ago

Multiply each number of the sequence

1, 2, 3 · · · by 3 , add 2 then write the

resulting numbers as a sequence

b) What is the tenth term of this sequence?

c) At what position 32comes in this

sequence ?

d) How many terms are there below 100 in

this sequence?

ans pls
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Answers

Answered by Alcaa
10

(a) The tenth term of this sequence is 32.

(b) The position of 32 comes in this  sequence at 10th place because we have calculated above that the tenth term of the sequence is 32.

(c) There are 32 terms in this sequence that are below 100.

Step-by-step explanation:

We are given the following sequence;

1, 2, 3, 4,..................

Now, it is specified that Multiply each number of the sequence  by 3 and then add 2 to the sequence, we get;

(1 \times 3) + 2, (2 \times 3) + 2, (3 \times 3) + 2, (4 \times 3) + 2,..............

5, 8, 11, 15, 18,..........., and so on.

We can see clearly that the resulting sequence is an A.P., where;

First term = a = 5

Common difference = d = 3

(a) The tenth term of this sequence is given by;

             a_n = a + (n -1) d  

             a_1_0 = 5 + (10 -1) \times 3

                    = 5 + 27 = 32

So, the tenth term of this sequence is 32.

(b) The position of 32 comes in this  sequence at 10th place because we have calculated above that the tenth term of the sequence is 32.

(c) Now, we have to find the number of terms that are there below 100 in  this sequence.

Firstly, as we know that the last term in this sequence will be 98 which is less than 100. So, we will find that at what position 98 comes in this sequence, that is ;

                   a_n = a + (n -1) d  

                    98 = 5 + (n -1) \times 3

                    98 -5= 3n-3

                    3(n-1) = 93

                    n-1 = \frac{93}{3}

                    n = 31 + 1 = 32

So, there are 32 terms in this sequence that are below 100.

Answered by DreamHacker
1

Answer

a) 5,8,11,14,17...

b) 32

c) 10th term

d) 32 term

Step-by-step explanation:

a) Sequence 1,2,3.....

   (1*3)+2,(2*3)+2,......

b)      an =  a + (n-1) d

       a10 = 5 + (10-1) *8

              = 5 + 27

              = 32

c)       an = a + (n-1) d

         98 = 5 + (n-1) * 3

    98 - 5 = 3n - 3

    3 (n-1) = 93

        n-1  = 93/3

            n = 31+1

               = 32

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