Question

in an ap sum of first term ,third term and fivth term is 39 and sum of second,fourth and sixth term is 51 find 20 term​

Answer 1

According to the question case 1=

( a + d) + (a + 3d) + (a + 5d) = 51

a+d+a+3d +a+5d=51

3a+9d=51

a + 3d = 17 ................[1]

similarily from case 2

a + 2d = 13 ..................[ 2 ]

Equation [ 1 ] - [ 2 ] :

d = 4

Find a:

Sub d = 4  into equation 2

a + 2(4) = 13

a + 8 = 13

a = 5

Form a nth term:

an = a1 + (n - 1)d

an = 5 + (n - 1)4

an = 5 + 4n - 4

an = 4n +1

Find the 20th term:

an = 4n + 1

a20 = 4(20) + 1

a20 = 80+1

a20=81

Answer 2

a + a + 2d + a + 4d = 39

a + d + a + 3d + a + 5d = 51

3a + 6d = 39

3a + 9d = 51

3a+6d=39

-3a-9d=-51

- 3d =-12

d = -12 / -3 ( - × - will cancel )

d = 4

3a + 6×4 = 39

3a + 24 = 39

3a = 39 - 24

3a = 15

a = 15 / 3

a = 5

an = a + ( n - 1 ) × d

= 5 + ( 20 - 1 ) × 4

= 5 + ( 80 - 4 )

= 5+ 76

= 81

a20 = 81

checking :

subtitute the values of a and d

a + 19 × d

5 + 19 × 4 = 81

5 + 76 = 81

81 = 81

thanku