The 4th and 7th terms of a G.P. are `1/(27)a n d1/(729)` respectively. Find the sum of `n` terms of the G.P.
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If the first, fifth and last terms of an `A.P.` is `l`, `m`, `p`, respectively, and sum of the `A.P.` is `((l+p)(4p+m-5l))/(k(m-l))` then `k` is
Correct Answer - A `(a)` Let common difference `=d` and number of terms `=n` `:.T_(s)=m=l+4dimpliesd=(m-l)//4` `:.T_(n)=p=l+((n-1)(m-1))/(4)` `implies n=((4P+m-5)/(m-1))` `:.` Sum of `n` terms of `A.P.=(n)/(2)["First term"+"Last term"]` `=[(4p+m-5l)/(m-l)]*(1)/(2)[l+p]`........`(i)` Comparing equation `(i)`...
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If the sum of the first `100` terms of an `AP` is `-1` and the sum of even terms lying in first `100` terms is `1`, then which of the following is not
Correct Answer - D `(d)` `x_(1)+x_(2)+x_(3)+……+x_(100)=(100)/(2)(x_(1)+x_(100))=-1` `impliesx_(1)+x_(100)=-(1)/(50)` `x_(2)+x_(4)+…+x_(100)=(50)/(2)(x_(1)+d+x_(100))=1` `impliesx_(1)+x_(100)+d=(1)/(25)` `impliesd=(3)/(50)` ltbr `x_(1)+x_(1)+99d=(-1)/(50)` `impliesx_(1)=(-149)/(50)` `x_(100)=x_(1)+99d` `=(-149)/(50)+99xx(3)/(50)` `=(74)/(25)`
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In `(3 3+1/(3 3))^n` if the ratio of 7th term from the beginning to the 7th term from the end is 1/6, then find the value of `ndot`
Correct Answer - B `(b)` `T_(r+1)=^(n)C_(r )a^(n-r)*b^(r )` where `a=2^(1//3)` and `b=3^(-1//3)` `T_(7)` from beginning `=^(9)C_(6)a^(n-6)b^(6)` and `T_(7)` from end `=^(n)C_(6)b^(n-6)a^(6)` `implies(a^(n-12))/(b^(n-12))=(1)/(6)` `implies2^((n-12)/(3))*3^((n-12)/(3))=6^(-1)` `n-12=-3impliesn=9`
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Write the following in exponential form and redical form. (i) Cube root of 729 (ii) Fifth root of 1990 (iii) Square root of 2001 (iv) Seventh root of
Correct Answer - (i) `(729)^(1/3), root(3)(729)` (ii) `(1990)^(1/5), root(5)(1990)` (iii) `(2001)^(1/2), sqrt(2001)` (iv) `(777)^(1/7), root(7)(777)`
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Find the sum of each of the following Aps: (i) 2, 7, 12, 17, … to 19 terms (ii) 9, 7, 5, 3,…. To 14 terms. (iii) -37, -33, -29,… to 12 terms (iv) `(1)
Correct Answer - (i) 893 (ii) -56 (iii) - 80 (iv) `(33)/(20)` (v) 5505
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(i) The sum of the first n terms of an AP is `((5n^(2))/(2) + (3n)/(2))`. Find the nth term and the 20th term of this AP. (ii) The sum of the first n
Correct Answer - `(i) T_(n) = (5n-1), T_(20) = 99 (ii) T_(n) = (3n +1), T_(25) = 76`
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अभाज्य गुणनखंड विधि से निम्न संख्याओ का वर्गमूल ज्ञात कीजिए : `{:((i)729,(ii)400,(iii)1764,(iv)4096),((v)7744,(vi)9604,(vii)5929,(viii)9216),((ix)529,
Correct Answer - `{:((i)27,(ii)20,(iii)42,(iv)64,(v)88),((vii)77,(viii)96,(ix)23,(x)90,):}`
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An AP 5, 8, 11…has 40 terms. Find the last term. Also find the sum of the last 10 terms.
First Term of the AP (a) = 5 Common difference (d) = 8 - 5 = 3 Last term = a40 = a + (40 - 1) d = 5 + 39 × 3...
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The volume of a cube is 729 cm^3 ,then its side is
Correct option is (B) 9 cm Let the side of cube be a. \(\therefore\) \(a^3=729=9^3\) \((\because\) Volume of cube \(=a^3)\) \(\Rightarrow\) a = 9 cm Hence, side of cube is 9 cm.
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The volume of a cube is 729 cm^3 ,then its side is
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