Question

Thefree numbers which are consecutive terms of an ap is 21 if the second number is reduced by one while the third is increased by 1 the number formed with three consecutive terms of a GP find these numbers.

Answer 1

Answer:

ANSWER

Let three numbers a,a+d,a+2d in A.P. then

a+a+d+a+2d=21 (Given)

3(a+d)=21⇒a+d=7. ......(1)

a=7−d ......(2)

Now, as per the given condition,

a,a+d−1,a+2d+1 are in G.P. then

So,

a

(a+d−1)

=

(a+d−1)

(a+2d+1)

36=a(8+d)

a

2

−15a+36=0 [On substituting value of d]

a=

2×1

−(−15)±

(−15)

2

−4×1×36

On solving, we get

a=12,3

so if a=12 then d=−5 [using equation(1)]

and similarly

if a=3 then d=4

We have two solutions of the numbers of the original A.P

(both solutions sharing the second term):

Solution i)12,7,2

Solution ii)3,7,11

Step-by-step explanation:

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