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IMG_20180830_131233_931.JPG sir question 3and 5 If an=sin(n pi)/6 then the value of sigma from n=1 to 6 a^2n If a1 =2, a2 =3+ al and an = 2 an _ 1 + 5 for n > 1, then t value of a20 is (a) 65 (c) 87 (a) 130 (b) 75 (d) 97 (b) 160

@ Roshni Afrin

$a_2=2a_1+5=9\\a_3=2a_2+5=23\\similarly\ a_4=51\\a_5=107\\\sum_{r=2}^{5}a_r=180$

IMG_20190124_233406.jpg EXANII'LE 21 The Of deviation a, a + d, a + 21, a + (2n 1) d, a + 2nd about the tncan is (a) Ans. (b) (b) (c) SOLUTION The mean of the given series is a + nd. (See chapter 52). Mean deviation about mean 1 I (a + rd) (a + nd) I (211 + 1) 2n d n 11 (11 + 1) x 2 211 + 1
@Shyam Both of the summation terms will be equal as | r - n | gives same values in interval [1,n] & [n,2n] so you can write the summation as double of summation from 1 to n.

Let $a_{1},a_{2},...........,a_{10}$  be a G.P. If  $\frac{a_{3}}{a_{1}}=25,$ then   $\frac{a_{9}}{a_{5}}$    equals :

• Option 1)

$4(5^{2})$

• Option 2)

$5^{4}$

• Option 3)

$2(5^{2})$

• Option 4)

$5^{3}$

Geometric Progession (GP) - A progression of non - zero terms, in which every term bears to the preceding term a constant ratio. - wherein eg 2, 4, 8, 16,- - - - - - and 100, 10, 1, 1/10,- - - - - - -     General term of a GP -   - wherein first term common ratio   Let first term of G.D be a and common ratio = r Now, Option 1)  Option 2)  Option 3)  Option 4)

If 19th term of a non-zero A.P. is zero, then its ( 49th term ) : ( 29th term ) is :

• Option 1)

2:1

• Option 2)

4:1

• Option 3)

3:1

• Option 4)

1:3

General term of an A.P. - - wherein First term number of term common difference Option 1)2:1Option 2)4:1Option 3)3:1Option 4)1:3

Let x,y be positive real numbers and m,n positive integers. The maximum value of the expression $\frac{x^{m}y^{n}}{(1+x^{2m})(1+y^{2n})}$  is :

• Option 1)

1

• Option 2)

$\frac{1}{2}$

• Option 3)

$\frac{1}{4}$

• Option 4)

$\frac{m+n}{6mn}$

Relation between AM, GM and HM of two positive numbers - - wherein Inequality of the three given means.    1 Option 1)  1Option 2)Option 3)  Option 4)

The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is $\frac{27}{19}.$

Then the common ratio of this series is :

• Option 1)

$\frac{4}{9}$

• Option 2)

$\frac{1}{3}$

• Option 3)

$\frac{2}{9}$

• Option 4)

$\frac{2}{3}$

Sum of infinite terms of a GP - - wherein first term common ratio   Lrt first term of G.P is a and the common ratio is r (r<1) Given,    2)  Given,   2)  From (1) and (2) since r<1  Option 1)  Option 2)  Option 3)  Option 4)

If the sum of the first 15 terms of the series $\left ( \frac{3}{4} \right )^{3}+\left ( 1\frac{1}{2} \right )^{3}+\left ( 2\frac{1}{4} \right )^{3}+3^{3}+\left ( 3\frac{3}{4} \right )^{3}+\cdots$

is equal to 225 k, then k is equal to :

• Option 1)

$9$

• Option 2)

$27$

• Option 3)

$108$

• Option 4)

$54$

Summation of series of natural numbers -   - wherein Sum of first n natural numbers     Summation of series of natural numbers - - wherein        Option 1)  Option 2)  Option 3)  Option 4)

If $_{}^{n}\textrm{C}_{4}$ , $_{}^{n}\textrm{C}_{5}$ , and $_{}^{n}\textrm{C}_{6}$  are in A.P., then n can be :

• Option 1)

$12$

• Option 2)

$9$

• Option 3)

$14$

• Option 4)

$11$

Arithmetic mean of two numbers (AM) - - wherein It is to be noted that the sequence a, A, b, is in AP where, a and b are the two numbers.  Option 1)Option 2)Option 3)Option 4)

Let $a_1, a_2, a_3, ..., a_{10}$ be in GP with $a_i > 0$ for $i = 1,2,..., 10$ and S be the set of pairs (r, k), r, k$\in N$( the set of natural numbers) for which

$\begin{vmatrix} \log_e a_1^ra_2^k & \log_e a_2^ra_3^k &\log_e a_3^ra_4^k \\ \log_e a_4^ra_5^k & \log_e a_5^ra_6^k & \log_e a_6^ra_7^k \\ \log_e a_7^ra_8^k &\log_e a_8^ra_9^k &\log_e a_9^ra_{10}^k \end{vmatrix} = 0$

Then the number of elements in S, is:

• Option 1)

4

• Option 2)

Infinitely many

• Option 3)

10

• Option 4)

2

Geometric Progession (GP) - A progression of non - zero terms, in which every term bears to the preceding term a constant ratio. - wherein eg 2, 4, 8, 16,- - - - - - and 100, 10, 1, 1/10,- - - - - - -   General term of a GP -   - wherein first term common ratio Apply coloumn operation  we get D = 0 OR    are in G.P. assume  Since  are in G.P. with common ratio 1  So,  Value of D become...

The sum of all two digit positive numbers which when divided by 7 yeild 2 or5 as remainder is  :

• Option 1)

1256

• Option 2)

1465

• Option 3)

1365

• Option 4)

1356

Summation of series of natural numbers -   - wherein Sum of first n natural numbers From the concept , General term will be  total = 1356Option 1)1256Option 2)1465Option 3)1365Option 4)1356

Let $S_{k}=\frac{1+2+3+\cdots +k}{k}.$  If ${S_{1}}^{2}+{S_{2}}^{2}+\cdots +{S_{10}}^{2}=\frac{5}{12}A,$  then A is equal to :

• Option 1)

$283$

• Option 2)

$301$

• Option 3)

$156$

• Option 4)

$303$

Summation of series of natural numbers -   - wherein Sum of first n natural numbers Since     Option 1)Option 2)Option 3)Option 4)

The product of three consecutive terms of a G.P. is 512. If 4 is added to each of the first and the second of these terms, the three terms now form an A.P. Then the sum of the original three terms of the given G.P. is :

• Option 1)

$24$

• Option 2)

$32$

• Option 3)

$36$

• Option 4)

$28$

Selection of terms in G.P. - If we have to take three terms in GP, we take them as - wherein Extension : If we have to take (2K+1) term in GP, we take them as     General term of an A.P. - - wherein First term number of term common difference    be three terms. Putting  Numbers are 4,8,16 or 16, 8,4  Sum of numbers = 4+ 8 + 16 = 28Option 1)Option 2)Option 3)Option 4)

Let a,b and c be the 7th,11th and 13th terms respectively of a non-constant A.P.If these are also the three consecutive terms of a G.P., then a/c is equal to :

• Option 1)

2

• Option 2)

7/13

• Option 3)

1/2

• Option 4)

4

General term of an A.P. - - wherein First term number of term common difference       Selection of terms in G.P. - If we have to take three terms in GP, we take them as - wherein Extension : If we have to take (2K+1) term in GP, we take them as From the concept  and  and  are in G.P    Option 1)  2Option 2)  7/13Option 3)  1/2Option 4)  4

Let S be the set of all triangles in the xy-plane,each having one vertex at the origin and the other two vertices  lie on coordinate axes with integral coordinates. If each triangles is S has area 50 sq. units, then the number of the elements in the set S is:

• Option 1)

9

• Option 2)

32

• Option 3)

18

• Option 4)

36

Number of Divisions - The number of divisors of a natural number   is  - wherein Where a1, a2 ....... are distinct prime and non negative integers.     Let  and  be vectors of  Area of  le is  Number of triangles   Option 1)  9Option 2)  32Option 3)  18Option 4)  36

The sum of the following series

$1+6+\frac{9(1^{2}+2^{2}+3^{2})}{7}+\frac{12(1^{2}+2^{2}+3^{2}+4^{2})}{9}+\frac{15(1^{2}+2^{2}+3^{2}+.....+5^{2}}{11})+.......$

up to 15 terms, is:

• Option 1)

7830

• Option 2)

7820

• Option 3)

7510

• Option 4)

7520

Summation of series of natural numbers -   - wherein Sum of first n natural numbers     Summation of series of natural numbers -   - wherein Sum of  squares of first n natural numbers     Summation of series of natural numbers - - wherein   The general term of the sequence will be Option 1)  7830Option 2)  7820Option 3)  7510Option 4)  7520

Let $a_1, a_2,... ....., a_{30}$ be an AP, $S = \sum_{i = 1}^{30}a_i$ and  $T = \sum_{i = 1}^{15}a_{(2i-1)}$ .

If $a_5 = 27$ and $S-2T = 75$

then $a_{10}$ is equal to:

• Option 1)

52

• Option 2)

57

• Option 3)

47

• Option 4)

42

Sum of n terms of an AP -   and Sum of n terms of an AP   - wherein first term common difference number of terms   Last term of an A.P.(l) - - wherein If a series has n terms last term is   Summation of terms of an AP       or,  Now,            Now use, Give  Now,                  Option 1) 52Option 2)57Option 3)47Option 4)42

If a, b and c be three distinct real numbers in G.P and $a+b+c = xb$ then $x$ cannot be:

• Option 1)

-2

• Option 2)

-3

• Option 3)

4

• Option 4)

2

General term of a GP -   - wherein first term common ratio   from the concept we have learnt  Let the three terms be  given    from the AM-GM         So,   So, Option 1)-2Option 2)-3Option 3)4Option 4)2
As we can see, the general term of the sequence is  Hence The sum : Now, putting n = 15, we get
Please, help me to get out of this problematic question... If a/b=2/3 and b/c=4/5, then find the value of (a+b)/(b+c)
@AMANDEEP   If a/b=2/3 and b/c=4/5, then find the value of (a+b)/(b+c)
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