There are 4n+ 1 terms in a sequence of which first 2n + 1 are in Arithmetic Progression and last 2n + 1 are in
Geometric Progression the common difference of Arithmetic Progression is 2 and common ratio of Geometric
Progression is 12. The middle term of the Arithuretic Progression is equal to middle term of Geometric Progression
Let midde tem of the sequence is Tm and Tm is the sum of infinite Geometric Progression whose sum of first two terms is (5/4)^2n and ratio of these terms is 9/16
On the basis of above information, answer the following questions
Number of terms in the given sequence is equal to -
(A9
(B) 17
(D) none
(C) 13
2.
Middle term of the guen sequence, ie T is equal to -
(A) 16,7
(B) 32/7
(C) 48/7
(D) 16/9
3.
(D) 48/7
First team of given sequence is equal to -
(A) 8/7.-2017
(B) -36/7
(C) 36/7
4.
Middle term of given A P. is equal to -
(A) 67
(B) 10/7
(C) 7817
(D) 11
5.
(C) 3
(D) 6
Sum of the terms of given A P is equal to -
(A) 6/7
(B) 7
Commrehension-2
m. be positive rational numbers, then
Answers
Answer:
There are 4n+1 terms in a sequence of which first 2n+1 are in Arithmetic Progression and last 2n+1 are in Geometric Progression in which the common difference of Arithmetic Progression is 2 and common ratio of Geometric Progression is
2
1
. The middle term of the Arithmetic Progression is equal to middle term of Geometric Progression. Let middle term of the sequence is T
m
and T
m
is the sum of infinite Geometric Progression whose sum of first Two terms is (
4
5
)
2
n and ratio of these terms is
16
9
.
First term of given sequence is equal to
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Answer
Correct option is
B
−
7
36
Given that, there are 4n+1 terms in a sequence of which first 2n+1 are in Arithmetic Progression and last 2n+1 are in Geometric Progression the common difference of Arithmetic Progression is 2 and common ratio of Geometric Progression is
2
1
.
Let, a be the first term of AP.
⇒ first term of GP is a+(2n+1−1)d=a+4n
Middle terms of AP and GP are equal.
⇒a+2n=
2
n
a+4n
-----(1)
But, Middle term of the whole sequence is T
m
which is sum of infinite GeometricProgression whose sum of first Two terms is (
4
5
)
2
n and ratio of these terms is
16
9
let A be the first term of infinite geometric series.
⇒A(1+r)=
16
25
n
⇒A=n
⇒T
m
=a+4n=
1−r
A
=
7
16n
-----(2)
from (1) and (2)
7
16n
−2n=
7.2
n
16n
⇒n=3
⇒T
m
=a+4n=
7
16n
=
7
48
.
∴ First term of the sequence a=
7
48
−12=
7
−36
Hence, option B.