30. Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A. P. and the ratio of 7th and (m – 1)th numbers is 5 : 9. Find the value of m.
30. Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A. P. and the ratio of 7th and (m – 1)th numbers is 5 : 9. Find the value of m.
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1 Answer
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30. Let A1, A2, A3 . Am be the m terms such that,
1, A1, A2, A3, . Am, 31 is an A.P.
So, a=1, first term of A.P
n=m+2, no. of term of A.P
l=31, last term of A.P. or (m+2)th term.
a+ [ (m+2) –1]d =31
1 + [m+1]d =31
(m+1)d =31 – 1=30
d =
The ratio of 7th and (m – 1) number is
? a1term=a
a2 =A1=a+d
a3 =A2=a+2d
:
a8 = A7 = a + 7d
9 [a+7d]=5 [a+ (m – 1)d]
9a+63d=5a+5 (m – 1)d.
9a – 5a=5 (m – 1)d – 63d.
4a= [5 (m – 1) –63]d.
So, putting a=1 and d=
4 × 1= [5 (m – 1) –63] ×
4 (m+1)= [5 (m – 1) –63] × 30
4m+4= [5m – 5 – 63] × 3
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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
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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