M.C.Q (1 Marks)

MCQ 2011 Mark

Determine the A.P. whose $3^{\text {rd }}$ term is 5 and the $7^{\text {th }}$ term is 9 .

A
$1,3,5,7,9, \ldots$
✓
$3,4,5,6,7, \ldots$
C
$-1,2,5,8, \ldots$
D
None of these

Answer

Correct option: B.

$3,4,5,6,7, \ldots$

(b) : We have,
$
\begin{aligned}
a_3 & =a+(3-1) d=a+2 d=5\ldots(i) \\
\text { and } a_7 & =a+(7-1) d=a+6 d=9\ldots(ii)
\end{aligned}
$
Solving the pair of linear equations (1) and (2), we get $a=3, d=1$
Hence, the required A.P. is $3,4,5,6,7, \ldots \ldots$

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MCQ 2021 Mark

If the $10^{\text {th }}$ term of an A.P. is 52 and $17^{\text {th }}$ term is 20 more than the $13^{\text {th }}$ term, then find the A.P.

A
$40,45,50, \ldots$.
B
$45,50,55, \ldots \ldots$
C
$17,22,27, \ldots$.
✓
$7,12,17, \ldots \ldots$

Answer

Correct option: D.

$7,12,17, \ldots \ldots$

(d) : We have, $a_{10}=52$
$
\Rightarrow a+9 d=52\ldots(i)
$
Also, $a_{17}=20+a_{13}$
$
\Rightarrow a+16 d=20+a+12 d \Rightarrow 4 d=20 \Rightarrow d=5
$
From (i), $a+9 \times 5=52 \Rightarrow a+45=52 \Rightarrow a=7$
Thus, the A.P. is $7,12,17, \ldots$

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MCQ 2031 Mark

If the sum of first $n$ terms of an $A.P.$ is given by $S_n=5 n^2+3 n$, then find its $n^{\text {th }}$ term.

✓
$10 n-2$
B
$10 n+3$
C
$2 n-10$
D
$2 n+10$

Answer

Correct option: A.

$10 n-2$

It is given that $S_n=5 n^2+3 n$
We know that $S_n-S_{n-1}=a_n$
$\Rightarrow 5 n^2+3 n-\left\{5(n-1)^2+3(n-1)\right\}=a_n$
$\Rightarrow 5 n^2+3 n-\left\{5\left(n^2+1-2 n\right)+3 n-3\right\}=a_n$
$\Rightarrow 5 n^2+3 n-\left\{5 n^2-7 n+2\right\}=a_n$
$\Rightarrow 10 n-2=a_n$

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MCQ 2041 Mark

If the sum of first $m$ terms of an $A.P$. is $2 m^2+3 m$, then what is its second term?

✓
$9$
B
$10$
C
$11$
D
$12$

Answer

Correct option: A.

$9$

We have, $S_m=2 m^2+3 m$
$\therefore S_1=2(1)^2+3(1)=5$ and
$S_2=2(2)^2+3(2)$
$=8+6=14$
Now, $a_1=S_1=5$ and $a_2=S_2-S_1$
$=14-5=9$
$\therefore $ Second term of the $A.P$. is $9$ .

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MCQ 2051 Mark

In an A.P., $a=-10, n=6$ and $a_n=10$, then the value of $d$ is
Riya and Diya said the following :
Riya : Common difference $=4$
Diya : Which of them is/are correct?

✓
Only Riya
B
Only Diya
C
Both Riya and Diya
D
Neither of them

Answer

Correct option: A.

Only Riya

(a) : Given, $a=-10, n=6, a_n=10$
Since, $a_n=a+(n-1) d$
$
\Rightarrow 10=-10+(6-1) d \Rightarrow 20=5 d \Rightarrow d=4
$

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MCQ 2061 Mark

The value of $a_{30}-a_{20}$ for the A.P. $2,7,12,17, \ldots$ is

A
100
B
10
✓
50
D
20

Answer

Correct option: C.

(c) : Given A.P. is $2,7,12,17, \ldots .$.
Here, $a=2, d=7-2=5$
Since $n^{\text {th }}$ term of an A.P. is $a_n=a+(n-1) d$
$
\therefore a_{30}=a+29 d=2+29(5)=147
$
and $a_{20}=a+19 d=2+19(5)=97$
Now, $a_{30}-a_{20}=147-97=50$

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MCQ 2071 Mark

The next term of the $A.P. \sqrt{18}, \sqrt{50}, \sqrt{95}, \ldots$ is

A
$\sqrt{146}$
B
$\sqrt{128}$
✓
$\sqrt{162}$
D
$\sqrt{200}$

Answer

Correct option: C.

$\sqrt{162}$

Common difference $(d)$ of $A.P. \sqrt{18}, \sqrt{50}$, $\sqrt{95}, \ldots$, is given by
$\sqrt{50}-\sqrt{18}=\sqrt{2 \times 25}-\sqrt{2 \times 9}=5 \sqrt{2}-3 \sqrt{2}=2 \sqrt{2}$
$\therefore \text { Fourth term, } a_4=a+3 d=\sqrt{18}+3 \times(2 \sqrt{2})$
$=3 \sqrt{2}+6 \sqrt{2}=9 \sqrt{2}$
$=\sqrt{2 \times 81}$
$=\sqrt{162}$

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MCQ 2081 Mark

If $k, 2 k-1$ and $2 k+1$ are three consecutive terms of an $A.P.,$ then the value of $k$ is

A
$2$
✓
$3$
C
$-3$
D
$5$

Answer

Correct option: B.

$3$

We have, $k, 2 k-1$ and $2 k+1$ are three consecutive terms of an $A.P.$
$\therefore (2 k-1)-(k)$
$=(2 k+1)-(2 k-1)$
$\Rightarrow k-1=2$
$\Rightarrow k=3$

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MCQ 2091 Mark

The common difference of the A.P. $\frac{1}{3 q}, \frac{1-6 q}{3 q}, \frac{1-12 q}{3 q}, \ldots .$. is

A
$q$
B
$-q$
✓
-2
D
2

Answer

Correct option: C.

-2

(c) : The common difference of the A.P.
$\frac{1}{3 q}, \frac{1-6 q}{3 q}, \frac{1-12 q}{3 q}, \ldots . .$. , is given by
$
\frac{1-6 q}{3 q}-\frac{1}{3 q}=\frac{1-6 q-1}{3 q}=\frac{-6 q}{3 q}=-2
$

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MATHS M.C.Q (1 Marks)