Question

1. If 'P'th , 'q' th and 'r' th terms of an AP are a , b, c respectively, then
show that a (q-r) + b(r-p) + c(p-q) = 0.
2. The sum of three numbers in an AP is -6 and their product is 64. Find the numbers.
​

Answer 1

Step-by-step explanation:

Let a = first term of the AP.

and

Let d = common difference of the AP

Now

a = A+(p-1).d.......(1)

b = A+(q-1).d.......(2)

c = A+(r-1).d........(3)

Subtracting 2nd from 1st , 3rd from 2nd and 1st from 3rd we get

a-b = (p-q).d......(4)

b-c = (q-r).d........(5)

c-a = (r-p).d.......(6)

multiply 4,5,6 by c,a,b respectively we have

c.(a-b) = c.(p-q).d......(4)

a.(b-c) = a.(q-r).d........(5)

b.(c-a) = b.(r-p).d.......(6)

a(q-r).d+b(r-p).d+c(p-q).d = 0(a(q-r)+b(r-p)+c(p-q)).d = 0

Now since d is common difference it should be non zero

Hence

a(q-r)+b(r-p)+c(p-q)= 0

(2) Es type Se kar lena

let the consecutive terms be

a-d,a,a+d

according to question

a+d+a+a-d = 6

3a = 6

therefore a = 2

situation 2

(a+d)×a×(a-d) = -120 ----------equation 2

putting a = 3 in equation 2

(2+d)×2×(2-d) = -120

2(4 -d^2) = -120

8 - 2d^2 = -120

- 2d^2 = -128

d^2 = 64

therefore d = 8

therefore

number 1 = 10

number 2 = 2

number 3 = -6