If 2nd, 8th, 44th terms of A.P. are 1st, 2nd and 3rd terms respectively of G.P. and first term of A.P. is 1 then the sum of first 20 terms of A.P. is
If 2nd, 8th, 44th terms of A.P. are 1st, 2nd and 3rd terms respectively of G.P. and first term of A.P. is 1 then the sum of first 20 terms of A.P. is
Option 1 -
970
Option 2 -
916
Option 3 -
980
Option 4 -
990
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1 Answer
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Correct Option - 1
Detailed Solution: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
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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