if the sum of first n term of ap is 1/2(3n2+2n) then find it's a20.
Answers
Answer:
a20= 119
Step-by-step explanation:
We start from the basics and recall that quantities are in Arithmetical Progression (A.P.) when they increase or decrease by a common difference. The common difference is found by subtracting any term of the sequence/series from that which follows it.
Thus a, a+d. a+2d, …………..a + (n-1)d
is an A.P. in which
the first term = a
and the common difference = d (a+d-a = a+2d - a-d = d)
Now,
Sum to 1st term = a
Sum to 2nd term = a + a+d = 2a+d
Sum to 3rd term = a + a+d + a+2d = 3a+3d and so on.
We observe that,
Sum to 2nd term - Sum to 1st term = 2a+d -a
= a+d = 2nd term ……………………………………………………………………………………………..…….(1)
Sum to 3rd term - Sum to 2nd term = 3a+3d - (2a+d) = 3a+3d - 2a-d
= a+2d = 3rd term
and so on.
By hypothesis,
Sum of the first n terms = 3n² + 2n
Putting n=1, sum to 1st term= 3.1² + 2.1 = 3+2 = 5 = first term
Putting n=2, sum to second term = 3.2² + 2.2 = 3.4 + 4 = 12 + 4 = 16
Putting n=3, sum of first three terms = 3.3² + 2.3 = 3.9 + 6 = 27 + 6 = 33
Putting n=4, sum of first four terms = 3.4² + 2.4 = 3.16 + 8 = 48 + 8 = 56
and so on.
∴ Using (1),
2nd term
= sum of first two terms - sum of first term
= 16 - 5 = 11 (Answer)
∴ Common difference = 11 - 5 = 6
and
The sequence is 5, 11, 17, 23……
a=5
d=6
a20= a+(n-1) d= 5+19×6= 119
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Answer:
a20= 119
Step-by-step explanation:
Thus a, a+d. a+2d, …………..a + (n-1)d
……
a=5
d=6
a20= a+(n-1) d= 5+19×6= 119
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