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^{6}- 12x^{5}+ ax^{4}+ bx^{3}+ cx^{2}+ dx + 64 = 0 has positive roots then find a,b,c,d.^{n }+b^{n }/a^{n-1}+b^{n-1 }is a G.M. between a and b, then the value of n is?^{2}, then the sum of squares of these n terms is_{1},a_{2},a_{3},.... be in harmonic progression with a_{1}= 5 and a_{20}= 25. The least positive integer n for which a_{n}< 0 is_{3}= h_{2}= 8 and x_{8}= h_{7}= 20, then x_{5}.h_{10}equals(A) 2560 (B) 2650 (C) 3200 (D) 1600

^{2}(1+b^{2}) + b^{2}(1+c^{2}) +c^{2}(1+a^{2}) > 6abc^{2 }+ cda/b^{2}+ dab/c^{2}+ abc/d^{2}> a+b+c+d_{r}be the r^{th}term of a sequence for r = 1,2,3,..... such that and T_{r}> 0 N. If sum of first n terms of this sequence is and maximum value of T_{5}is where p & q are coprime numbers then value of (p + q) is-If y is the(i)betweenH.M.andx, prove that,zIf harmonic mean of two numbers is1/(y-x) + 1/(y-z) = 1/x + 1/z(ii)

, their arithmetic mean4and the geometric mean'A'satisfy the relationG. Find the numbers. [2A + G^{2}= 27Answer :]6, 3In order to save time I have given two questions in one post. Please give answer of both the questions ans please donot give any link.

THANK YOUa.1/6(n

^{2}+3n+8)b.n/6(n

^{2}+3n+8)c.1/6(n

^{2}-3n+8)d.n/6(n

^{2}-3n+8)what to do in this question how to solve pllz give me hint, method of difference we can use here??

ncards numbered from 1 ton. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 1224.If the smaller of the numbers on the removed cards is

k, thenk– 20 =Please

explain the highlighted stepin the following solution:

a.(log l - log a)/log r

b.l - (log l - log a)/log r

c.(log a - log l)/log r

d.l + (log l - log a)/log r

_{r}be the r^{th}term of a sequence for r = 1,2,3,..... such that and T_{r}> 0 N. If sum of first n terms of this sequence is and maximum value of T_{5}is where p & q are coprime numbers then value of (p + q) is- (1) 3381 (2) 3268 (3) 3496 (4) 3167 [

Answer :]40/9 km/hNO LINKS PLEASE$\mathbf{129}\mathbf{.}Sumtontermsoftheseries\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+.....,is\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(a\right)\frac{{n}^{3}}{3\left(n+1\right)\left(n+2\right)\left(n+3\right)}\left(b\right)\frac{{n}^{3}+6{n}^{2}-3n}{6\left(n+2\right)\left(n+3\right)\left(n+4\right)}\phantom{\rule{0ex}{0ex}}\left(c\right)\frac{15{n}^{2}+7n}{4n\left(n+1\right)\left(n+5\right)}\left(d\right)\frac{{n}^{3}+6{n}^{2}+11n}{18\left(n+1\right)\left(n+2\right)\left(n+3\right)}$

Q). The least value of $a\in R$ for which $4a{x}^{2}+\frac{1}{x}\ge 1$ for all $x>0$, is

(a) $\frac{1}{64}$

(b) $\frac{1}{32}$

(c) $\frac{1}{27}$

(d) $\frac{1}{25}$

$\mathbf{4}\mathbf{.}\mathbf{3}\mathbf{}Ifforsomepositiveintegern1,wehave1+x+{x}^{2}+.....+{x}^{n-1}=0,then\sum _{k=0}^{n}\left({x}^{k}+\frac{1}{{x}^{k}}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(A\right)-2\left(B\right)-1\left(C\right)1\left(D\right)2$

term of ap^{th} , q^{th}and the r^{th}beH.P.respectively, prove thatx, y, zyz(q-r) + zx(r-p) + xy(p-q) = 0.

NO LINK PLEASE

Q). A pack contains n cards numbered from 1 to n. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the removed cards is k, then $k-20=$ ?

x−y+1=0 and 7x−y−5=0. If its

diagonals intersect at (−1, −2), then

which one of the following is a vertex of

this rhombus ?

Q). ${a}_{1},{a}_{2},{a}_{3},.......,{a}_{n}\to A.P$ positive terms, then

$\frac{{a}_{1}+{a}_{2n}}{\sqrt{{a}_{1}}+\sqrt{{a}_{2}}}+\frac{{a}_{2}+{a}_{2n-1}}{\sqrt{{a}_{2}}{\displaystyle +}{\displaystyle \sqrt{{a}_{3}}}}+............+\frac{{a}_{n}+{a}_{n+1}}{\sqrt{{a}_{n}}{\displaystyle +}{\displaystyle {{\displaystyle \sqrt{a}}}_{n+1}}}$

^{3})]>3^{3}a^{2}b^{2}