Find the sum to `n` terms of the series: `1+5+12+22+35+`
1 Answers
Talk Doctor Online in Bissoy App
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)`
2 Answers 1 views
The sum of the series `1+(9)/(4)+(36)/(9)+(100)/(16)+…` infinite terms is
Correct Answer - A `(a)` The given series can be written as `1^(3)+(1^(3)+2^(3))/(1+3)+(1^(3)+2^(3)+3^(3))/(1+3+5)` `t_(n)=(1^(3)+2^(3)+3^(3)+.....n^(3))/(1+3+5+...+(2n-1))` `=(n^(2)(n+1)^(2))/(4n^(2))=((n+1)^(2))/(4)` `=(1)/(4)(n^(2)+2n+1)` `:.S_(n)=(1)/(4)[sum_(k=1)^(n)k^(2)+2sum_(k=1)^(n)k+n]` `:.S_(n)=(1)/(4)[(n(n+1)(2n+1))/(6)+n(n+1)+n]` `:.S_(16)=(1)/(4)[(16.17.33)/(6)+16.17+16]` `=(1)/(4)[88xx17+16xx17+16]=446`
2 Answers 1 views
The sum of `n` terms of series `ab+(a+1)(b+1)+(a+2)(b+2)+…+(a+(n-1)(b+(n-1))` if `ab=(1)/(6)` and `(1+b)=(1)/(3)` is
Correct Answer - C `(c )` `S=ab+[ab+(a+b)+1]+[ab+2(a+b)+2^(2)]+…+[ab+(n-1)(a+b)+(n-1)^(2)]` `=nab+(a+b)sum_(r=1)^(n-1)r+sum_(r=1)^(n-1)r^(2)` `=nab+(a+b)(n(n-1))/(2)+((n-1)(n)(2n-1))/(6)` `=(n)/(6)[1+(n-1){1+2n-1}]` `=(n)/(6)[1+2n(n-1)]=(n)/(6)(1-2n+2n^(2))`
2 Answers 1 views
The absolute value of the sum of first `20` terms of series, if `S_(n)=(n+1)/(2)` and `(T_(n-1))/(T_(n))=(1)/(n^(2))-1`, where `n` is odd, given `S_(n
Correct Answer - B `(b)` We have `S_(n)-S_(n-2)=T_(n)+T_(n-1)` (Taking `n` to be odd) `:.T_(n-1)(1+(T_(n))/(T_(n-1)))=((n+1)/(2))-((n-1))/(2)=1` (As `S_(n)=(n+1)/(2)`, if `n` is odd) `:.T_(n-1)(1+(n^(2))/(1-n^(2)))=1` `:.T_(n-1)=-(n^(2)-1)` when `n` is odd Also, `S_(m)=S_(m-1)+T_(m)` If `m` is...
2 Answers 1 views
9. Write notes on: (a) Continuous series (b) Discrete series (c) Individual series
(a) Continuous series is the third series followed by individual and discrete series. (b) In these series all the items are divided in certain groups, but these groups are not continuous, therefore...
2 Answers 1 views
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.
Correct Answer - `3/2(1-1/3^(n))`
2 Answers 1 views
Find the sum of `n` terms of the series whose `n t h` term is: `2n^2-3n+5`
Correct Answer - `(n)/(6)(4n^(2)-3n+23)`
2 Answers 1 views
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
2 Answers 1 views
(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`
2 Answers 1 views
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...
2 Answers 1 views