the
In
an
arithmetic Sequence, Sum of
first 9 terms is
279 and the sum of first
20 terms is
1280,
then,
a) what is the 5th term of this sequence?
b)what is 16th term of the
Sequence ?
c) write
down
the sequence ? n²+n
Answers
Answer:
dm on @__princely___ for algebra doubts
Step-by-step explanation:
Sn = n ( a1 + an ) / 2
In this case:
S9 = 9 ( a1 + a9 ) / 2 = 279
S20 = 20 ( a1 + a20 ) / 2 = 1280
9 ( a1 + a9 ) / 2 = 279 Multiply both sides by 2
9 ( a1 + a9 ) = 279 * 2
9 a1 + 9 a9 = 558
On the other side:
an = a1 ( n - 1 ) d
in this case: n = 9, n - 1 = 8
a9 = a1 + 8 d
9 a1 + 9 a9 = 558
9 a1 + 9 ( a1 + 8 d ) = 558
9 a1 + 9 a1 + 9 * 8 d = 558
9 a1 + 9 a1 + 72 d = 558
18 a1 + 72 d = 558
20 ( a1 + a20 ) / 2 = 1280 Multiply both sides by 2
20 ( a1 + a20 ) = 1280 * 2
20 a1 + 20 a20 = 2560
an = a1 ( n - 1 ) d
In this case: n = 20, n - 1 = 19
a20 = 20 a1 + 19 d
20 a1 + 20 a20 = 2560
20 a1 + 20 ( a1 + 19 d ) = 2560
20 a1 + 20 a1 + 20 * 19 d = 2560
20 a1 + 20 a1 + 380 d = 2560
40 a1 + 380 d = 2560
Now you must solve system of 2 equations with 2 unknows:
18 a1 + 72 d = 558
40 a1 + 380 d = 2560
The solutions are :
a1 = 7
d = 6
an = a1 + ( n - 1 ) d
a5 = 7 + ( 5 - 1 ) * 6
a5 = 7 + 4 * 6
a5 = 7 + 24
a5 = 31
an = a1 + ( n - 1 ) d
a16 = 7 + ( 16 - 1 ) * 6
a16 = 7 + 15 * 6
a16 = 7 + 90
a16 = 97
Your sequence:
7, 13, 19 , 25, 31, 37, 43, 49, 55, 61, 67, 73, 79, 85, 91, 97, 103, 109, 115, 121