Question

If 18, a, b, -3 are in A.P. Then find a + b.

Answer 1

Hint: To solve this problem, we should know the properties of Arithmetic Progressions. Arithmetic progressions have an initial term and a common difference which is added to the previous term to get the next term. The general arithmetic progression of ‘n’ terms can be written as $\text{x, x+d, x+2d, x+3d,}.........\text{x+}\left( \text{n-1} \right)\text{d}$ where x- first term; d- Common difference. We have to correlate the above general expansion with the given A.P in the question to find a, b and finally a + b.

Complete step-by-step answer:
 The general arithmetic progression of ‘n’ terms can be written as $\text{x, x+d, x+2d, x+3d,}.........\text{x+}\left( \text{n-1} \right)\text{d}$ where x- first term; d- Common difference.
Given A.P in the question is 18, a, b, -3. Comparing this with the general arithmetic progression gives,
First term = x = 18.$\to (1)$
Second term = x + d = a $\to (2)$
Third term = x + 2d = b $\to (3)$
Fourth term = x+3d = -3 $\to (4)$
We should find the common difference which is d from the equations (1) and (4).
(4) – (1) gives
x + 3d – x = -3-18
3d = -21
d = -7
Using the value of x and din equations (2) and (3) we get
Second term =\[x+d=18+\left( -7 \right)=18-7=11=a\]
Third term = \[x+2d=18+2\left( -7 \right)=18-14=4=b\]
The required answer is a + b which is equal to $a+b=11+4=15$
$\therefore $$a+b=15$

Note:There is another way to do this problem. The sum of a + b from the general terms is given by$a+b=x+d+x+2d=2x+3d$. The term $2x+3d$ can also be written as $2x+3d=\left( x \right)+\left( x+3d \right)$ which according to the equations (1) and (4) is equal to $2x+3d=\left( x \right)+\left( x+3d \right)=18+\left( -3 \right)=18-3=15$and it is same as the answer we got above. In general, we can say that the sum of the first and fourth terms of an A.P will be equal to the second and third terms of the same A.P.