Sequences And Series(11th Grade > Mathematics ) Questions and Answers
Explanation:-
Answer: Option C. -> 4:
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
Explanation:-
Answer: Option A. -> 7:
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.........
Explanation:-
Answer: Option C. -> Rs. 20500:
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.
Explanation:-
Answer: Option C. -> n + 2−n - 1:
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.
Explanation:-
Answer: Option D. -> G2 = AH:
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
Explanation:-
Answer: Option B. -> √mn:
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.