u6+u7=39,u29=87,jumlah 45 suku pertama

u6+u7=39
a+5b+a+6b=39
2a+11b=39

u29=87
a+28b=87

mencari nilai a dan b dengan metode eliminasi substitusi
2a+11b=39 | x1 | 2a+11b=39
a+28b=87 | x2 | 2a+56b= 174
~> -45b= -135
-> b=3
a= 3

S45=45/2.(2.3+(45-1).3)
= 45/2(6+44.3)
=45/2(138)
=45.69
=3105

U6 + U7 = 39
a + 5b + a + 6b = 39
2a + 11b = 39 ... pers 1

U29 = 87
a + 28b = 87 ... pers 2

pers 1 & 2
2a + 11b = 39
a + 28b = 87 |×2
maka
2a + 11b = 39
2a + 56b = 174
-----------------------(-)
- 45b = - 135
b = 3

subtitusi ke pers
2a + 11b = 39
2a + 11(3) = 39
2a + 33 = 39
2a = 39 - 33
a = 6 ÷ 2
a = 3

S45 = ½ • n ( 2a + ( n - 1)b)
= ½ • 45 ( 2•3 + ( 45 - 1)3)
= 45/2 ( 6 + 132)
= 45/2 • 138
= 3.105