Two terms of an arithmetic sequence are a7 = 32 and a19 =140. B. Find n such that Sn = 511

n = 60.22

Step-by-step explanation:

Hello

To find Sn, we need to draw out equations for each a₇ and a₁₉

In an arithmetic progression,

Sn = a + (n-1)d

Where Sn = sum of the A.P

a = first term

d = common difference

a₇ = 32

32 = a + (7-1)d

32 = a + 6d equation (i)

a₁₉ = 140

140 = a + (19-1)d

140 = a + 18d equation (ii)

Solve equation (i) and (ii) simultaneously

From equation (i)

32 = a + 6d

Make a the subject of formula

a = 32 - 6d equation (iii)

Put equation (iii) into equation (ii)

140 = (32 - 6d) + 18d

140 = 32 - 6d + 18d

Collect like terms

140 - 32 = 12d

12d = 108

d = 108 / 12

d = 9

Put d = 9 in equation (i)

32 = a + 6(9)

32 = a + 54

a = 32 - 54

a = -22

When Sn = 511

Sn = a + (n - 1)d

Substitute and solve for n

511 = -22 + (n-1) × 9

511 = -22 + 9n - 9

511 = -31 + 9n

511 + 31 = 9n

542 = 9n

n = 542 / 9

n = 60.22

2500

step-by-step explanation:

your answer is the frist one

hop i