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Home » The sum of three numbers which are consecutive terms of an A.P. is 21. If the second number is reduced by 1 and the third is increased by 1, we obtain three consecutive terms of a G.P. Find the numbers.
The sum of three numbers which are consecutive terms of an A.P. is 21. If the second number is reduced by 1 and the third is increased by 1, we obtain three consecutive terms of a G.P. Find the numbers.
CBSE | Class 11 | Exercise 20.5 | Geometric Progressions | Maths | RD Sharma
The sum of three numbers which are consecutive terms of an A.P. is 21. If the second number is reduced by 1 and the third is increased by 1, we obtain three consecutive terms of a G.P. Find the numbers.

 Solution:

Consider the first term of an A.P. to be ‘a’ and let its common difference be‘d’.

Therefore, we can write: a1 + a2 + a3 = 21

Where the three numbers are as follows:

a, a + d, and a + 2d

So, we get:

3a + 3d = 21

Or, a + d = 7.

We have: d = 7 – a …. (i)

Now, we have:

a, a + d – 1, and a + 2d + 1

All these are in geometric progression

\left( a+d-1 \right)/a\text{ }=\text{ }\left( a+2d+1 \right)/\left( a+d-1 \right)

{{\left( a\text{ }+\text{ }d-1 \right)}^{2}}~=~a\left( a\text{ }+\text{ }2d\text{ }+\text{ }1 \right)

{{a}^{2}}~+\text{ }{{d}^{2}}~+\text{ }1\text{ }+\text{ }2ad-2d-2a\text{ }=\text{ }{{a}^{2}}~+\text{ }a\text{ }+\text{ }2da

{{\left( 7-a \right)}^{2}}-3a\text{ }+\text{ }1-2\left( 7-a \right)\text{ }=\text{ }0

49\text{ }+\text{ }{{a}^{2}}-14a-3a\text{ }+\text{ }1-14\text{ }+\text{ }2a\text{ }=\text{ }0

{{a}^{2}}-15a\text{ }+\text{ }36\text{ }=\text{ }0

{{a}^{2}}-12a-3a\text{ }+\text{ }36\text{ }=\text{ }0

a\left( a-12 \right)-3\left( a-12 \right)\text{ }=\text{ }0

a\text{ }=\text{ }3\text{ }or\text{ }a\text{ }=\text{ }12

d\text{ }=\text{ }7-a

d\text{ }=\text{ }7-3\text{ }or\text{ }d\text{ }=\text{ }7-12

d\text{ }=\text{ }4\text{ }or-5

Then,

For a = 3 and d = 4, the A.P is 3, 7, 11

For a = 12 and d = -5, the A.P is 12, 7, 2

∴ The numbers are 3, 7, 11 or 12, 7, 2

i

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