Question

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​

Answer 1

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