If 3, a, b, c are in A.P. and 3, a – 1, b + 1 are in G.P. Then arithmetic mean of a, b and c is
If 3, a, b, c are in A.P. and 3, a – 1, b + 1 are in G.P. Then arithmetic mean of a, b and c is
Option 1 -
11
Option 2 -
10
Option 3 -
9
Option 4 -
13
-
1 Answer
-
Correct Option - 1
Detailed Solution:3, a, b, c are in A.P.
a – 3 = b – a (common diff.)
2a = b + 3
and 3, a – 1, b + 1 are in G.P.
a2 + 1 – 2a = 3b + 3
a2 – 8a + 7 = 0 &n
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Similar Questions for you
First term = a
Common difference = d
Given: a + 5d = 2 . (1)
Product (P) = (a1a5a4) = a (a + 4d) (a + 3d)
Using (1)
P = (2 – 5d) (2 – d) (2 – 2d)
-> = (2 – 5d) (2 –d) (– 2) + (2 – 5d) (2 – 2d) (– 1) + (– 5) (2 – d) (2 – 2d)
= –2 [ (d – 2) (5d – 2) + (d – 1) (5d – 2) + (d – 1) (5d – 2) + 5 (d – 1) (d – 2)]
= –2 [15d2 – 34d + 16]
at
-> d = 1.6
a, ar, ar2, ….ar63
a+ar+ar2 +….+ar63 = 7 [a + ar2 + ar4 +.+ar62]
1 + r = 7
r = 6
S20 = [2a + 19d] = 790
2a + 19d = 79 . (1)
2a + 9d = 29 . (2)
from (1) and (2) a = –8, d = 5
= 405 – 10
= 395
3, 7, 11, 15, 19, 23, 27, . 403 = AP1
2, 5, 8, 11, 14, 17, 20, 23, . 401 = AP2
so common terms A.P.
11, 23, 35, ., 395
->395 = 11 + (n – 1) 12
->395 – 11 = 12 (n – 1)
32 = n – 1
n = 33
Sum =
=
= 6699
a + d, a + 7d and a + 43d are 1st, 2nd, 3rd term of G.P.
⇒ (a + 7d)2 = (a + d) (a + 43d)
⇒ a2 + 49d2 + 14d = a2 + 44ad + 43d3
⇒ 6d2 = 30ad
⇒ d2 = 5d
⇒ d = 0, 5
a = 1, d = 5
= 10 [95 + 2]
= 970
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