Question

3 Consider the arithmetic sequence 6,10,14
(a)
Find its first term?
(b) Find its common difference?
Write its algebraic form
(d) Write its 10 term
(e) Is 2020 is include in that sequence ? Justify your answer?
Is the difference of two terms of that sequence 2020?
Justify your answer?
Find sum of its first 10 terms?
(h) Write algebraic expression for the sum of n terms of
the sequence?
(1) Which is the first 3 digit number in that sequence ?
6) How many two digits numbers are there in the sequence?​

Answer 1

$Given \: Arithmetic \: sequence : 6,10,14$

$a)First \:term (a = a_{1} ) = 6$

$b) Common \:term (d) = a_{2} - a_{1}$

$= 10 - 6$

$= 4$

$c) n^{th}\: term \: (a_{n})= a + (n-1)d$

$= 6 + ( n - 1 ) \times 4$

$= 6 + 4n - 4$

$= 4n + 2$

$d) a = 6 , d = 4 \: and \: n = 10$

$10^{th} \:term = a_{10}$

$= a + 9d$

$= 6 + 9 \times 4$

$= 6 + 36$

$= 42$

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Answer 2

Solution

a) First term (a = a1) = 6

b) Common term (d) = a2 - a1

=> 10-6

=> 4

c) nth term = a + (n - 1)d

=> 6 + (n - 1) × 4

=> 6 + 4n - 4

=> 4n + 2

d) a = 6, d = 4 & n = 10

10th term = a10

=> a + 9d

=> 6 + 9 × 4

=> 6 + 36

=> 42

hope it helps ! ☺️