Asked by Balaji

In an arithmatic sequence, sum of 9 terms is 279 and sum of 20 terms is 1280.
Find 5th term ?
Find 16th term ?
Write the sequence.
Answered by Bosnian
Sum of a certain number of terms of an arithmetic sequence:

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






Answered by Balaji
S9 = 279
F5 = 279/9 = 31
S20 = 1280
S20-S9 = 1280-279 =1001
=[sum of F11 to F20]
F15 =1001/11 = 91.
F15-F5 = 91-31 = 60 = 10 d
d = 60/10 = 6
F16 = 91 + 6 =97
F1 =31-4d =31-24= 7
so, 7,13,19,25,..........
There are no AI answers yet. The ability to request AI answers is coming soon!