11th Grade > Mathematics
SEQUENCES AND SERIES MCQs
:
C
Let there be 2n terms in the given G.P. with first term a and the common ratio r.
Then, a r2−1(r−1) = 5a (r2−1)(r2−1) ⇒ r + 1 = 5 ⇒ r = 4
:
D
S=a(1−rn)1−rP=anrn(n−1)2R=1a+1ar+1ar2+⋯⋯+1arn−1=1a(1−1rn)1−1r=1arn−1(rn−1r−1)p2=a2nrn(n−1)(SR)=a2nrn(n−1)
:
A
(b, c, d) Given xn = xn+1 √2
∴ x1 = x2√2, x2 = x3√2, xn = xn+1√2
On multiplying x1 = xn+1 (√2)n ⇒ xn+1 = x1(√2)n
Hence xn = x1(√2)n−1
Area of Sn = x2n = x2n2n−1 < 1 ⇒ 2n−1 > x21 (x1 = 10)
∴ 2n−1 > 100
But 27 > 100, 28>100, etc.
∴ n - 1 = 7, 8, 9....... ⇒ n = 8, 9, 10.........
:
C
It will take 10 years for harikiran to pay off Rs. 10000 in 10 yearly
installments.
∵ He pays 10% annual interest on interest on remaining amount
∴ Money given in first year
= 1000 + 10000×10100 = Rs. 2000
Money given in second year = 1000 + interest of (10000 - 1000) with interest rate 10% per annum
= 1000 + 9000×10100 = Rs. 1900
Money paid in third year = Rs. 1800 etc.
So money given by Jairam in 10 years will be Rs. 2000, Rs.1900, Rs. 1800, Rs. 1700,......,
which is in arithmetic progression,
whose first term a = 2000 and d = -100
Total money given in 10 years = sum of 10 terms of arithmetic
progression
= 102[2(2000) + (10 - 1)(-100)] = Rs. 15500
Therefore, total money given by harikiran
= 5000 + 15500 = Rs. 20500.
:
D
Tn=1[n(n+1)2]=2[1n−1n+1]
put n=1,2,3,.........,(n+1)
T1=2[11−12],T2=2[12−13],...........,
Tn+1=2[1n+1−1n+2]
Hence sum of (n+1) terms = n+1∑k=1Tk
⇒ Sn+1=2[1−1n+2] ⇒ Sn+1=2(n+1)n+2
:
C
The sum of the first n terms is
Sn = (1 - 12) + (1 - 122) + (1 - 123) + (1 - 124 + ........... + (1 - 12n)
= n - { 12 + 122 +............+ 12n}
= n - 12(1−12n1−12) = n - (1 - 12n) = n - 1 + 2−n.
Trick: Check for n = 1, 2 i.e., S1 = 12, S2 = 54 and
(c) ⇒ S1 = 12 and S2 = 2 + 2−2 - 1 = 54.
:
D
Let A = a+b2, G =√ab and H = 2aba+b.
Then G2 = ab ..................(i)
and AH = (a+b2).2aba+b = ab ................(ii)
From (i) and (ii), we have G2 = AH
:
B
Given that
ap+q=arp+q−1=m and ap−q=arp−q−1=n.
⇒m×n=arp+q−1 x arp−q−1
=a2r2(p−1)
=(arp−1)2
⇒√mn=arp−1=ap
Thus, the pth term of the GP is √mn.
Aliter: Each term in a G.P. is the geometric mean of the terms equidistant from it. Here, (p+q)th and (p−q)th terms are equidistant from the pth term.
∴ pth term (√mn) will be G.M. of (p+q)th and (p−q)th terms.
:
B
(
11 -
12) + (
12 -
13) + (
13 -
14) + .........+ (
1n -
1n+1)
= 1 -
1n+1 =
nn+1
:
A
π∑i=1 = 3π∑i=1i - 2 π∑i=11 = 3
n(n+1)2 - 2n =
n(3n−1)2