three num
bers are in an ap and their sum is 21 if the first and second are decreased by1 and third is increased by 7 they form a G.P find the three numbers in an Ap
Answers
Answer:
Write sentences in your notebook to illustrate the following aspects of the possessive forms of the nouns.
Answer:
Let the first term of an A.P. be ‘a’ and its common difference be‘d’.
a1 + a2 + a3 = 21
Where, the three number are: a, a + d, and a + 2d
So,
3a + 3d = 21 or
a + d = 7.
d = 7 – a …. (i)
Now, according to the question:
a, a + d – 1, and a + 2d + 1
they are now in GP, that is:
(a + d - 1)/a = (a + 2d + 1)/(a + d - 1)
(a + d – 1)2 = a(a + 2d + 1)
a2 + d2 + 1 + 2ad – 2d – 2a = a2 + a + 2da
(7 – a)2 – 3a + 1 – 2(7 – a) = 0
49 + a2 – 14a – 3a + 1 – 14 + 2a = 0
a2 – 15a + 36 = 0
a2 – 12a – 3a + 36 = 0
a(a – 12) – 3(a – 12) = 0
a = 3 or a = 12
d = 7 – a
d = 7 – 3 or d = 7 – 12
d = 4 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