Question

the sum of first 9 terms of an arithmetic sequence is 279 and the sum of first 20 terms is 1280.find the 5 th and 16 th term ?write the sequence.​

Answer 1

Answer:

$sn = \binom{n}{2}(2a + (n - 1)d)$

S9= (9/2) (2a+(9-1)d

=

$\frac{9}{2} \: 2 (a + 4d)$

$9a + 36d = 279$

a+4d=

$\frac{279}{9}$

a+4d=31----(i)

$s20 = \frac{n}{2} (2a + (n - 1)d$

$\frac{20}{2} (2a + (20- 1)d)$

=10(2a+19d)=1280

2a+19d=128----(ii)

From (i) and (ii),

a +4d =31 (×2)

2a+19d=128

»2a+19d=128 (-)

2a+8d=62

=11d= 66

d= 66/11

d=6

a=31-4d

= 31-4(6)

31-24

=7

5th term

a+(n-1)d

a+5d

From eqn (i),

a+4d= 31

5th term is 31

16th term

a+15d

7+ 15(6)

7+90

=97

16 terms is 97

Sequence/AP= a,a+d, a+2d....

7,13,19,25,31,37.........

Answer 2

Answer:

Please any one prove this equation correctly

Step-by-step explanation:

Very hard to find this