Maths M.C.Q (1 Marks)

A.P. is $\frac{1}{3},\frac{1-3\text{b}}{3},\frac{1-6\text{b}}{3}, .....\ \text{is}$
Common difference $\text{d}=\frac{1-3\text{b}}{3}-\frac{1}{3}=\frac{1-6\text{b}}{3}-\frac{(1-3\text{b})}{3}$
$\frac{1-3\text{b}-1}{3}=\frac{1-6\text{b}-1+3\text{b}}{3}$
$-3\text{b}=-3\text{b}$
$-\text{b}=-\text{b}$

Given, $A.P. 3, 7, 11, 15, .....$
$S_n= 406$
First term, $a = 3$
and Difference, $d = 7 - 3 = 4$
We know, $\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\ 406=\frac{\text{n}}{2}[2(3)+(\text{n}-1)4]$
$⇒ 406 × 2 = n[6 + 4n - 4]$
$⇒ 812 = n[4n + 2]$
$⇒ 812 = n × 2[2n + 1]$
$⇒ 406 = n[2n + 1]$
$ \Rightarrow 406=2 n^2+n $
$ \Rightarrow 2 n^2+n-406=0 $
$ \Rightarrow 2 n^2+29 n-28 n-406=0$
$\Rightarrow n(2n + 29) - 14(2n + 29) = 0$
$\Rightarrow (n - 14)(2n + 29) = 0$
Here,
$n - 14 = 0$
$\Rightarrow n = 14$
and $2n + 29 = 0$
$2n = -29$
$\text{n}=\frac{-29}{2}$
We know term never ne negative
So, number of terms is $14$
Hence, correct choice is $(D)$.

In the given problem, we are given first, second and last term of an A.P. We need to find its sum.
So, here
First term $= a$
Second term $(a_2) = b$
Last term $(l) = 2a$
Now, using the formula $a_n= a + (n - 1)d$
$a_2= a + (2 - 1)d$
$b = a + d$
$d = b - a .....(i)$
Also,
$a_n= a + (n - 1)d$
$2a = a + nd - d$
$2a - a = nd - d$
$\frac{\text{a}+\text{d}}{\text{d}}=\text{n}\ .....\text{(ii)} $
Furthere as we know,
$\text{S}_\text{n}=\frac{\text{n}}{2}[\text{a}+\text{l}]$
Substituting (ii) in the above equation, we get
Using (i), we get
$\text{S}_\text{n}=\frac{\text{a}+(\text{b}-\text{a})}{2(\text{b}-\text{a})}(3\text{a})$
$\text{S}_\text{n}=\frac{\text{b}}{2(\text{b}-\text{a})}(3\text{a})$
Thus,
$\text{S}_\text{n}=\frac{3\text{ab}}{2(\text{b}-\text{a})}$
Therefore, the correct option is $(C)$.

Given,
First term, $a$
Last term, $l$
Sum of $l$ terms, $S_l$
and Difference, $\text{d}=\frac{\text{l}^2-\text{a}^2}{\text{k}-(\text{l}+\text{a})}\ .....(\text{i})$
We know, $l = a + (n - 1)d$
put in eq. (i) $\text{d}=\frac{(\text{a}+(\text{n}-1)\text{d})^2-\text{a}^2}{\text{k}-[(\text{a}+(\text{n}-1)\text{d})+\text{a}]}$
$\Rightarrow\ \text{d}=\frac{\text{a}^2+((\text{n}-1)\text{d})^2+2\text{a}(\text{n}-1)\text{d}-\text{a}^2}{\text{k}-[\text{a}+\text{nd}-\text{d}+\text{a}]}$
$\text{d}=\frac{(\text{n}-1)^2\text{d}^2+2\text{a}(\text{n}-1)\text{d}}{\text{k}-[2\text{a}+(\text{n}-1)\text{d}]}$
$\Rightarrow\ \text{d}=\frac{\text{d}[(\text{n}-1)^2\text{d}+2\text{a}(\text{n}-1)]}{\text{k}-[2\text{a}+(\text{n}-1)\text{d}]}$
$\Rightarrow k - [2a+ (n - 1)d] = (n - 1)2 d + 2a (n - 1)$
$\Rightarrow k = (n - 1)2 d + 2a (n - 1) + [2a + (n - 1)d]$
$\Rightarrow k = (n - 1)[(n - 1)d + 2a] + [2a + (n - 1)d]$
$\Rightarrow k = 2a + (n - 1)d [n - 1 + 1]$
$\Rightarrow k = (2a + (n - 1)d)n$
We know, $\text{S}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
Now divide by $2$ both side
$\Rightarrow\ \frac{\text{k}}{2}=(2\text{a}+(\text{n}-1)\text{d})\frac{\text{n}}{2}$
$\Rightarrow\ \frac{\text{k}}{2}=\text{S}$
$\Rightarrow\ \text{k}=2\text{S}$
Hence, correct chice is $(B)$.

Given,
$a_7= 34$
and $a_{13}= 64$
We know, $a_n= a + (n - 1)d$
$7^{th}$ term, $a_7= a + (7 - 1)d$
$\Rightarrow 34 = a + 6d .....(i)$
$13^{th}$ term, $a_{13}= a + (13 - 1)d$
$\Rightarrow 64 = a + 12d .....(ii)$
By subracting eq. $(i)$ from $(ii)$ from eq. $(ii)$
$\Rightarrow 64 - 34 = a + 21d - (a + 6d)$
$\Rightarrow 30 = a + 12d - a - 6d$
$\Rightarrow 30 = 6d$
$\Rightarrow\ \text{d}=\frac{30}{6}$
Put the value of d in eq. (i)
$\Rightarrow 34 = a + 6(5)$
$\Rightarrow 34 = a + 30$
$\Rightarrow a = 34 - 30 = 4$
Now we have to find $18^{th}$ term,
$\Rightarrow a_{18}= 4 + (18 - 1)5$
$\Rightarrow a_{18}= 4 + 17 \times 5$
$\Rightarrow a_{18}= 4 + 85$
$\Rightarrow a_{18}= 89$
Hence, correct choice is $(C)$.

$S_n$ is the sum of first n terms
Last term $n^{th}$ term $= S_n- S_{n-1}$

Given,
$S_1$ is the sum of an A.P. of $n$ odd number of terms,
$S_2$ is the sum of the terms if the series in odd plaves,
Let, A.P. is $a_2,a_2, a_3, ....., a_n$
Here,
$n$ is odd
$d$ is difference
We know, $\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d]}$
then, $\text{S}_1=\frac{\text{n}}{2}[2\text{a}_1+(\text{n}-1)\text{d}]$
Now, $S^2$ be the sum if the terms of the series in odd places.
Number of terms, $\frac{\text{n}+1}{2}$
Difference, $d = 2d$
We know, $\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\ \text{S}_2=\frac{\frac{\text{n}+1}{2}}{2}\bigg[2\text{a}_1+\Big(\frac{\text{n}+1}{2}-1\Big)2\text{d}\bigg]$
$\Rightarrow\ \text{S}_2=\frac{\text{n}+1}{2}\Big[2\text{a}_1+\frac{\text{n}-1}{2}\times2\text{d}\Big]$
$\Rightarrow\ \text{S}_2=\frac{\text{n}+1}{4}[2\text{a}_1+(\text{n}-1)\text{d}]$
Now, we have to find $\frac{\text{S}_1}{\text{S}_2}$,
$\Rightarrow\ \frac{\text{S}_1}{\text{S}_2}=\frac{\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]}{\frac{\text{n}+1}{4}[2\text{a}_1+(\text{n}-1)\text{d}]}$
$\Rightarrow\ \frac{\text{S}_1}{\text{S}_2}=\frac{4\text{n}}{2(\text{n}+1)}$
$\Rightarrow\ \frac{\text{S}_1}{\text{S}_2}=\frac{2\text{n}}{(\text{n}+1)}$
Hence, correct chice is $(A)$.

Given,
First term, $a = 1$
Last term, $a_n= 11 = l$
and Sum of $11$ term, $S_{11}= 36$
We know, Sum of n terms,
$\Rightarrow\ \text{S}_\text{n}=\frac{\text{n}}{2}(\text{a}+\text{l})$
$\Rightarrow\ 36=\frac{\text{n}}{2}(1+11)$
$\Rightarrow\ 36\times2=\text{n}\times12$
$\Rightarrow\ \text{n}=\frac{36\times2}{12}$
$\Rightarrow\ \text{n}=6$
Hence, correct choice is $(b)$.

Given,
$\frac{\text{Sum of A.P.}_1}{\text{Sum of A.P.}_2}=\frac{\text{S}_\text{n}}{\text{S}'_\text{n}}=\frac{5\text{n}+9}{9\text{n}+6}\ .....{\text{(i)}}$
We know,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
For $A.P._1$ $\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
For $A.P._2$ $\text{S}'_\text{n}=\frac{\text{n}}{2}[2\text{a}'+(\text{n}-1)\text{d}]$
Put the value in Eq. $(i)$
$\Rightarrow\ \frac{\text{S}_\text{n}}{\text{S}'_\text{n}}=\frac{\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]}{\frac{\text{n}}{2}[2\text{a}'+(\text{n}-1)\text{d}']}$
$\Rightarrow\ \frac{5\text{n}+9}{9\text{n}+6}=\frac{[2\text{a}+(\text{n}-1)\text{d}]}{[2\text{a}'+(\text{n}-1)\text{d}']}$
Now, put the $n = 2n - 1$
$\Rightarrow\ \frac{5(2\text{n}-1)+9}{9(2\text{n}-1)+6}=\frac{[2\text{a}+(2\text{n}-1-1)\text{d}]}{[2\text{a}'+(2\text{n}-1-1)\text{d}']}$
$\Rightarrow\ \frac{10\text{n}-5+9}{18\text{n}-9+6}=\frac{[2\text{a}+(2\text{n}-2)\text{d}]}{[2\text{a}'+(2\text{n}-2)\text{d}']}$
$\Rightarrow\ \frac{10\text{n}+4}{18\text{n}-3}=\frac{2\text{a}+2(\text{n}-1)\text{d}}{2\text{a}'+2(\text{n}-1)\text{d}'}$
$\Rightarrow\ \frac{2(5\text{n}+2)}{3(6\text{n}-1)}=\frac{2[\text{a}+(\text{n}-1)\text{d}]}{2[\text{a}'+(\text{n}-1)\text{d}']}$
$\Rightarrow\ \frac{2(5\text{n}+2)}{3(6\text{n}-1)}=\frac{[\text{a}+(\text{n}-1)\text{d}]}{[\text{a}'+(\text{n}-1)\text{d}']}$
We kmow, $a_n= a + (n - 1)d$
$\Rightarrow\ \frac{2(5\text{n}+2)}{3(6\text{n}-1)}=\frac{\text{a}_\text{n}}{\text{a}'_\text{n}}$
Now put $n = 18$
$\Rightarrow\ \frac{2[5(18)+2]}{3[6(18)-1]}=\frac{\text{a}_{18}}{\text{a}'_{18}}$
$\Rightarrow\ \frac{2[90+2]}{3[108-1]}=\frac{\text{a}_{18}}{\text{a}'_{18}}$
$\Rightarrow\ \frac{2\times92}{3\times107}=\frac{\text{a}_{18}}{\text{a}'_{18}}$
$\Rightarrow\ \frac{184}{321}=\frac{\text{a}_{18}}{\text{a}'_{18}}$
Hence, the correct option is $(D)$.

Sum of $5 + 9 + 13 + .....$ to n term
$=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
Here, $a = 5, d = 9 - 5 = 4$
$\therefore\ \text{Sum}=\frac{\text{n}}{2}[2\times5+(\text{n}-1)\times4]$
$=\frac{\text{n}}{2}[10+4\text{n}-4]$
$=\frac{\text{n}}{2}[6+4\text{n}=\text{n}(3+2\text{n})]$
and sum of $7 + 9 + 11 + .....$ to $(n + 1)$ terms
$=\frac{\text{n}+1}{2}[2\times7+(\text{n}+1-1)2]$
$=\frac{\text{n}+1}{2}[14+2\text{n}]=(\text{n}+1)(7+\text{n})$
$\therefore\ \frac{5+9+13+\ .....\text{to n terms}}{7+9+11+\ .....\text{ to }(\text{n}+1)\text{ terms}}=\frac{17}{16}$
$\Rightarrow\ \frac{\text{n}(3+2\text{n})}{(\text{n}+1)(7+\text{n})}=\frac{17}{16}$
$ \Rightarrow 16 n(3+2 n)=17(n+1)(7+n) $
$ \Rightarrow 48 n+32 n^2=17\left(n^2+8 n+7\right) $
$ \Rightarrow 48 n+32 n^2=17 n^2+136 n+119 $
$ \Rightarrow 48 n+32 n^2-17 n^2-136 n-119=0 $
$ \Rightarrow 15 n^2-88 n-119=0 $
$ \Rightarrow 15 n^2-105 n+17 n-119=0$
$ \because\ 15\times(-119)=1785 -1785 = 17\times(105) -88 = 17 - 105  $
$\Rightarrow 15n(n - 7) + 17(n - 7) = 0$
$\Rightarrow (n - 7)(15n + 17) = 0$
Either, $n - 7 = 0$, then $n = 7$ or $15n + 17 = 0,$ then $\text{n}=\frac{-17}{15}$ which is not possible being fraction.
$\therefore\ \text{n}=7$

Here, we are given four terms which are in A.P.,
First term $\left(a_1\right)=$
Second term $\left(a_2\right)=$
Third term $\left(a_3\right)=$
Fourth term $\left(\mathrm{a}_4\right)=$
So, in an A.P. the difference of two adjacent terms is always constant. So, we get,
$d=a_2-a_1$
$d=a-18....(i)$
Also,
$d=a_4-a_3$
$d = -3 - b .....(ii)$
Now, on equating $(1)$ and $(2)$, we get,
$a - 18 = -3 - b$
$a + b = 18 - 3$
$a + b = 15$
Therefore, $a + b = 15$
Hence the correct option is $(D)$.

In the given problem, we are given $\mathrm{S}_{\mathrm{p}}=\mathrm{q}$ and $\mathrm{S}_{\mathrm{q}}=\mathrm{p}$
We need to find $S_{p+q}$
Now, as we know,
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
So
$\text{S}_\text{p}=\frac{\text{p}}{2}[2\text{a}+(\text{p}-1)\text{d}]$
$\text{q}=\frac{\text{p}}{2}[2\text{a}+(\text{p}-1)\text{d}]$
$2\text{q}=2\text{ap}+\text{p}(\text{q}-1)\text{d}\ .....(\text{i})$
Similarly,
$\text{S}_\text{q}=\frac{\text{q}}{2}[2\text{a}+(\text{q}-1)\text{d}]$
$\text{p}=\frac{\text{q}}{2}[2\text{a}+(\text{q}-1)\text{d}]$
$2\text{p}=2\text{aq}+\text{q}(\text{q}-1)\text{d}\ .....\text{(ii)}$
Subtracting $(ii)$ from $(i)$, we get
$2q - 2p = 2ap + [p(p - 1)d] - 2aq - [q(q - 1)d]$
$2q - 2p = 2a(p - q) + [p(p - 1) -1(q - 1)]d$
$-2(p - q) = 2a(p - q) + [(p^2- q^2 - (p - q)]$
$-2 = 2a + (p + q - 1)d .....(iii)$
Now,
$\text{S}_{\text{p}+\text{q}}=\frac{\text{p}+\text{q}}{2}[2\text{a}+(\text{p}+\text{q}-1)\text{d}]$
$\text{S}_{\text{p}+\text{q}}=\frac{(\text{p}+\text{d})}{2}(-2)$
$\text{S}_{\text{p}+\text{q}}= -(\text{p}+\text{d})$
Thus, $S_{p+q}= -(p + q)$
Hence, the correct option is $(d)$.