27) The sum to n terms of an AP is 3n²-2n Find the 16th term of the AP
Answers
put n=1
S1=3(1)^2+2(1)
s1=3+2
s1=5
putn=2
s2=3(2)^2+2(2)
s2=12+4
s2=16
now,
a=t1=s1
a=t1=5
t2=s2-s1
=16-5
t2=11
a=5
d=t2
tn=5+6n-6
the nth term is 6n-1
Step-by-step explanation:
Given :-
The sum of n terms of an AP is 3n²-2n
To find :-
Find the 16th term of the AP ?
Solution :-
Given that
The sum of n terms of an AP = 3n²-2n
Sn = 3n²-2n -----(1)
Put n = 1 in (1) then
S1 = 3(1)²-2(1)
=> S1 = 3(1)-2
=> S1 = 3-2
=> S1 = 1
=> Sum of first terms = 1
=> First term = 1
and
Put n = 2 in (1) then
=>S2 = 3(2)²-2(2)
=> S2 = 3(4)-4
=> S2 = 12-4
=> S2 = 8
Sum of first two terms = 8
=> First term+Second term = 8
=> Second term = 8-First term
=> Second term = 8-1
=> Second term = 7
We have,
a1 = a = 1
a2 = 7
Common difference
= Second term - First term
=> Common difference = a2-a1
=> d = 7-1
=> d = 6
Now
We know that
nth term of an AP (an) = a+(n-1)d
We have,
a = 1
d = 6
n = 16 given
16th term of the AP = a16
=> a16 = 1+(16-1)(6)
=>a16 = 1+15(6)
=>a16 = 1+90
=>a16 = 91
Answer:-
16th term of the AP is 91
Used formulae:-
→ nth term of an AP = a+(n-1)d
- a = First term
- d = Common difference
- n = Number of terms