Question

A linear sequence starts a+3b a+7b a+11b the 2nd term has value 19 the 5th term has value 67 work out the values of a and b

Answer 1

Answer:

$a = -9$

$b = 4$

Step-by-step explanation:

Given

Sequence: a+3b, a+7b, a+11b

2nd term = 19

5th term = 67

Required

Find a and b

First, the 5th term needs to be calculated;

Using formula for Arithmetic Progression (AP), the formula goes thus

$T_n = T_1 + (n - 1)d$

Where n = 5

T_1 = a + 3b ------------ FIrst term

$d = T_2 - T_1 or T_3 - T_2$ --- Difference between two successive terms

$d = a + 7b - (a + 3b)$

$d = a + 7b -a - 3b$

$d = a - a + 7b - 3b$

$d = 4b$

So, $T_n = T_1 + (n - 1)d$ becomes

$T_5 = a + 3b + (5 - 1)4b$

$T_5 = a + 3b + (4)4b$

$T_5 = a + 3b + 16b$

$T_5 = a + 19b$

Now that we have values for 2nd and 5th term;

From the second, T2 = 19 and T5 = 67

This gives

$a + 7b = 19$ --- Equation 1

$a + 19b = 67$ ---- Equation 2

Make a the subject of formula in (1)

$a = 19 - 7b$

Substitute these values in equation 1

$a + 19b = 67$ becomes

$19 - 7b + 19b = 67$

$19 + 12b = 67$

Collect like terms

$12b= 67 - 19$

$12b = 48$

Divide both sides by 12

$\frac{12b}{12} = \frac{48}{12}$

$b = 4$

Recall that b = 4

Substitute a = 19 - 7b and nothing will hire

$a = 19 - 7(4)$

$a = 19 - 28$

$a = -9$

Hence, the values of a and b are -9 and 4 respectively.