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The greatest integer less than or equal to the sum of first 100 terms of the sequence 1 3 , 5 9 , 1 9 2 7 , 6 5 8 1 , . . . . . . . . . . . . i s     e q u a l     t o . . . . . . . . . . . .  

0 2 Views | Posted 3 weeks ago
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    Answered by

    Raj Pandey | Contributor-Level 9

    3 weeks ago

    t n = 3 n − 2 n 3 n = 1 − ( 2 3 ) n

    S 1 0 0 = 1 0 0 − 2 3 ( ( 2 3 ) 1 0 0 − 1 ) 2 3 − 1 = 1 0 0 + 2 . ( 2 1 0 0 − 3 1 0 0 ) 3 1 0 0

    = 1 0 0 − 2 + 2 1 0 1 3 1 0 0 = 9 8 + ( 2 3 ) 1 0 0 . 2 < 9 8 + 1

Similar Questions for you

1In an increasing arithmetic progression a1, a2,....an if a5 = 2 and product of a1, a5 and a4 is greatest, then the value of dis equal to
A
alok kumar singh

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)

d P d d = –2 [ (d – 2) (5d – 2) + (d – 1) (5d – 2) + (d – 1) (5d – 2) + 5 (d – 1) (d – 2)]

= –2 [15d2 – 34d + 16]

d = 8 5   o r   2 3

at  ( 8 5 ) , product attains maxima

-> d = 1.6

In a 64 terms GP if sum of total terms is seven times sum of odd terms, then common ratio is
A
alok kumar singh

a, ar, ar2, ….ar63

a+ar+ar2 +….+ar63 = 7 [a + ar2 + ar4 +.+ar62]

a ( 1 − r 6 4 ) ( 1 − r ) = 7 a ( 1 − r 6 4 ) ( 1 − r 2 )            

1 + r = 7

r = 6

In an arithmetic progression if sum of 20 terms is 790 and sum of 10 terms is 145, then S15 – S5 is (when Sn denotes sum of n terms)
A
alok kumar singh

S20 = 2 0 2 [2a + 19d] = 790

2a + 19d = 79              . (1)

S 1 0 1 0 2 [ 2 a + 9 d ] = 1 4 5

2a + 9d = 29                . (2)

from (1) and (2) a = –8, d = 5

S 1 5 − S 5 = 1 5 2 [ 2 a + 1 4 d ] − 5 2 [ 2 a + 4 d ]

= 1 5 2 [ − 1 6 + 7 0 ] − 5 2 [ − 1 6 + 2 0 ]  

= 405 – 10

= 395

If 3, 7, 11,...., 403 = AP1

2, 5, 8, ...., 401 = AP2

Find sum of common term of AP1 and AP2
A
alok kumar singh

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)

3 8 4 1 2 = n − 1    

32 = n – 1

n = 33

Sum =  3 3 2 [2×11+ (32)12]

=  3 3 2 [22 + 384]

= 6699

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
A
alok kumar singh

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.

a − 1 3 = b + 1 a − 1              

a2 + 1 – 2a = 3b + 3

a2 – 8a + 7 = 0                         &n

...more

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