Math, asked by alokjose00, 10 months ago

The first term of an arithmetic sequence is 10 and its common difference is 3

Write the first three terms of the sequence

Verify whether 100 is a term of this sequence?

a) Write its 10th term

b) Write its nth term ( Algebraic form)

c) Can the difference between any two terms of this sequence be 45?

d) Which term of this sequence is 181?​

Answers

Answered by chitluridevicharan
5

Step-by-step explanation:

1)The 1st 3 terms in an arthematic sequence are a,a+d,a+2d where a is 1st term and d is common difference.so the three terms are 10,10+3,10+2*3= 10,13,16.

2) nth term of arthematic sequence is a+(n-1)d

so let a+(n-1)d=100

10+(n-1)3=100

3n-3=90

3n=93

n=31

so we can say that 100 is the term of arthematic sequence.

3)nth term of arthematic sequence is a+(n-1)d

so nth term of this sequence is 10+3(n-1)

=10+3n-3

=7+3n

4)Yes. Because 45 is a multiple of 3. Like nd is a multiple of d.

5)Let a+(n-1)d=181

10+(n-1)3=181

7+3n=181

3n=174

n=58

So 181 is 58th term of the arthematic sequence.

Answered by amitnrw
1

Given : The first term of an arithmetic sequence is 10 and its common difference is 3

To Find :  first three terms of the sequence

whether 100 is a term of this sequence?

Write its 10th term

Write its nth term

Can the difference between any two terms of this sequence be 45

Which term of this sequence is 181

Solution:

a  = 10

d = 3

aₙ = a + (n -1)d

=> aₙ = 10 + (n -1)3

=> aₙ = 3n  + 7

nth term = 3n + 7

first three terms of the sequence

10 , 13 , 16  

3n  + 7 = 100

=> 3n = 93

=> n = 31 hence 31st  term  is 100

3n  + 7 = 181

=> 3n = 174

=> n = 58   58th term is  174

Difference between any 2 terms can be 3k

Hence 45 can be difference

10 + 45 = 55

3n + 7 = 55 => n = 16

Difference between 1st and 16th term  =  45  

Difference between nth and  n + 15 th term will always be 45

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