Question

If 18, a, b, -3 are in AP, then a+ b is

Answer 1

Answer:

Let d be the common difference

a - 18 = $d_{1}$

b - a = $d_{2}$

- 3 - b = $d_{3}$

As we know that the d is same throughout the AP

So we can say that

$d_{1}$ = $d_{2}$ = $d_{3}$

putting the values of $d_{1}$ & $d_{2}$

a - 18 = b - a

b = 2a - 1 8

now putting the value of $d_{2}$ & $d_{3}$

b - a = -3 -b

2b -a = 18 - 3

2( 2a - 18) - a = 15

4a - a = 15 + 36

3a = 51

a = 17

b = 2a - 18

b = 2(17) - 18

b = 34 -18

b = 16

finding a + b

= 17 + 16

= 33

Answer 2

Answer:

15

Step-by-step explanation:

Let p, be the first term  i.e; 18

And, d, be the common difference,

From given series,

4th term =p+3d=−3

⟹18+3d=−3

Hence,  d=−7

So, a=p+d=18−7=11

And,  b=p+2d=18−14=4

Therefore,

a+b=11+4=15