plss answer fast 10 the chapter arithmetic sequence
Answers
Step-by-step explanation:
Given :-
Second term of an arithmetic sequence is 9 and the tenth term is 41.
To find :-
Find the following :
I) Common difference
II) 25th term
Solution :-
We know that
If a is the first term and d is the common difference of an AP then nth term of an AP is
an = a+(n-1)d
Given that
Second term of an AP = 9
=> a2 = 9
=> a+(2-1)d = 9
=> a+(1)d = 9
=> a+d = 9 ------------(1)
Tenth term of the AP = 41
=> a10 = 41
=> a+(10-1)d = 41
=> a+9d = 41 --------(2)
On subtracting (1) from (2) then
a + 9d = 41
a + d = 9
(-) (-) (-)
__________
0 + 8d = 32
__________
=> 8d = 32
=> d = 32/8
=> d = 4
Common difference = 4
On substituting the value of d in (1) then
=> a+4 = 9
=> a = 9-4
=> a = 5
First term = 5
Now,
25th term of the AP = a25
=> a25 = 5 +(25-1)(4)
=> a25 = 5+(24)(4)
=> a25 = 5+96
=> a25 = 104
Therefore, 25th term = 104
Answer:-
The common difference of the AP = 4
25th term of the AP = 104
Check:-
If a = 5 and d = 4 then
a2 = a+d = 5+4 = 9
a10 = 5+9(4) = 5+36 = 41
Verified the given relations in the given problem.
Used formulae:-
→ nth term of an AP is an = a+(n-1)d
→ a = first term
→ d = common difference
→ n = number of terms