11th And 12th > Mathematics
SEQUENCE AND SERIES MCQs
:
D
Obviously, 7th term of corresponding A.P is 18 and 8th
term will be 17. a + 6d = 18 and a+7d = 17
Solving these, we get d = 156 and a = 156
Therefore 15th term of this A.P.
= 156+14×156 = 1556
Hence the required 15th term of the H.P. is 5615
:
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
:
B
214. 418. 8116. 16132 .........∞
= 214+28+316+........ = 2S, where S is given by
S = 14 + 2 18 + 3 116 +4 132 + ............∞ .........(i)
⇒ 12 S = 18 + 216 + 332 + 464 + ................∞ ..........(ii)
Subtracting (ii) from (i), we get S = 1.
Hence required product = 21 = 2.
:
A
Given series, let Sn = 1 + 25 + 352 + 453 + .............+ n5n−1
15Sn = 15 + 252 + 353 + ..............+ n5n
Subtracting,
(1 - 15)Sn = 1 + 15 + 152 + 153 + ............+ upto n terms - n5n
⇒ 45 Sn = 1−15n45 - n5n ⇒ Sn = 2516 - 4n+516×5n−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.
:
A
Let Sn be the sum of the given series to n terms, then
Sn = 1 + 2x + 3
x2 + 4
x3 + ....... + n
xn−1 ..........(i)
xSn = x + 2
x2 + 3
x2 + ...........+ n
xn ..........(ii)
Subtracting (ii) from (i), we get
(1-x)Sn = 1 + x +
x2 +
x3 + ......... to n terms -n
xn
= (
(1−xn)(1−x)) - nxn
⇒ Sn =
(1−xn)−nxn(1−x)(1−x)2 =
1−(n+1)xn+nxn+1(1−x)2