Question

Choose the correct alternative answer for each of the following sub questions.
(1) The sequence –10, –6, –2, 2,...

(A) is an A.P., Reason d= –16	(B) is an A.P., Reason d= 4
(C) is an A.P., Reason d= –4	(D) is not an A.P.

(2) First four terms of an A.P. are ....., whose first term is –2 and common difference is –2.

(A) –2, 0, 2, 4	(B) –2, 4, –8, 16
(C) –2, –4, –6, –8	(D) –2, –4, –8, –16

(3) What is the sum of the first 30 natural numbers ?

(A) 464

(B) 465

(C) 462

(D) 461

(4) For an given A.P. t₇ = 4, d = –4, n = 101, then a = ....

(A) 6

(B) 7

(C) 20

(D) 28

(5) For an given A.P. a = 3.5, d = 0, n = 101, then t_n = ....

(A) 0

(B) 3.5

(C) 103.5

(D) 14.5

(6) In an A.P. first two terms are –3, 4 then 21^st term is ...

(A) –143

(B) 143

(C) 137

(D) 17

(7) If for any A.P. d = 5 then t₁₈ – t₁₃ = ....

(A) 5

(B) 20

(C) 25

(D) 30

(8) Sum of first five multuiples of 3 is...

(A) 45

(B) 55

(C) 15

(D) 75

(9) 15, 10, 5,... In this A.P sum of first 10 terms is...

(A) –75

(B) –125

(C) 75

(D) 125

(10) In an A.P. 1^st term is 1 and the last term is 20. The sum of all terms is = 399 then n = ....

(A) 42

(B) 38

(C) 21

(D) 19

Answer 1

(1) The given sequence is –10, –6, –2, 2,...

Here,
First term (a) = a₁ = –10
Second term = a₂ = –6
Third term = a₃ = –2

Common difference (d) = a₂ – a₁ = –6 – (–10) = 4
= a₃ – a₂ = –2 – (–6) = 4

Since, a₂ – a₁ = a₃ – a₂

Thus, the given sequence is an A.P.

Hence, the correct option is (B).


(2) It is given that,
First term (a) = –2
Common difference (d) = –2

Second term = a + d = –2 + (–2) = –4
Third term = a + 2d = –2 + 2(–2) = –6
Fourth term = a + 3d = –2 + 3(–2) = –8

Thus, first four terms of the A.P. are –2, –4, –6, –8

Hence, the correct option is (C).

(3) The given series is 1 + 2 + 3 + ... + 30

Here,
a = 1
d = 1
n = 30

$$\begin{gathered} S_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right) \\ {S}_{30}=\frac{30}{2}\left(2a+\left(30-1\right)d\right) \\ =\frac{30}{2}\left(2\left(1\right)+29\left(1\right)\right) \\ =15\left(2+29\right) \\ =15\times 31 \\ =465 \end{gathered}$$

Hence, the correct option is (B).

(4) It is given that,
t₇ = 4
d = –4
n = 101

Now,
$$\begin{gathered} t_n=a+\left(n-1\right)d \\ {t}_7=a+\left(7-1\right)d \\ \Rightarrow 4=a+6\left(-4\right) \\ \Rightarrow 4=a-24 \\ \Rightarrow a=4+24 \\ \Rightarrow a=28 \end{gathered}$$

Hence, the correct option is (D).

(5) It is given that,
a = 3.5
d = 0
n = 101

$$\begin{gathered} t_n=a+\left(n-1\right)d \\ =3.5+\left(n-1\right)\left(0\right) \\ =3.5 \end{gathered}$$

Hence, the correct option is (B).

(6) It is given that,
a = –3
a₂ = 4

We know that,
$$\begin{gathered} a_2=a+\left(2-1\right)d \\ \Rightarrow 4=-3+d \\ \Rightarrow d=7 \end{gathered}$$

Now,
$$\begin{gathered} a_{21}=a+\left(21-1\right)d \\ =-3+20\left(7\right) \\ =-3+140 \\ =137 \end{gathered}$$

Hence, the correct option is (C).

(7) It is given that,
d = 5

Now,
$$\begin{gathered} t_n=a+\left(n-1\right)d \\ {t}_{18}-t_{13}=\left(a+\left(18-1\right)d\right)-\left(a+\left(13-1\right)d\right) \\ =\left(a+17d\right)-\left(a+12d\right) \\ =5d \\ =5\left(5\right) \\ =25 \end{gathered}$$

Hence, the correct option is (C).

(8) The given sequence is 3, 6, 9,...
Here,
a = 3
d = 3
n = 5

We know that,
$$\begin{gathered} S_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right) \\ {S}_5=\frac{5}{2}\left(2a+\left(5-1\right)d\right) \\ =\frac{5}{2}\left(2\left(3\right)+4\left(3\right)\right) \\ =\frac{5}{2}\left(6+12\right) \\ =\frac{5}{2}\left(18\right) \\ =5\left(9\right) \\ =45 \end{gathered}$$

Hence, the correct option is (A).

(9) The given sequence is 15, 10, 5,...
Here,
a = 15
d = –5

We know that,
$$\begin{gathered} S_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right) \\ {S}_{10}=\frac{10}{2}\left(2a+\left(10-1\right)d\right) \\ =5\left(2\left(15\right)+9\left(-5\right)\right) \\ =5\left(30-45\right) \\ =5\left(-15\right) \\ =-75 \end{gathered}$$

Hence, the correct option is (A).

(10) It is given that,
First term (a) = 1
Last term (t_n) = 20
Sum of terms (S_n) = 399

We know that,
$$\begin{gathered} t_n=a+\left(n-1\right)d \\ {S}_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right) \\ \Rightarrow S_n=\frac{n}{2}\left(a+\left(a+\left(n-1\right)d\right)\right) \\ \Rightarrow S_n=\frac{n}{2}\left(a+t_n\right) \\ \Rightarrow 399=\frac{n}{2}\left(1+20\right) \\ \Rightarrow 399=\frac{21n}{2} \\ \Rightarrow 21n=399\times 2 \\ \Rightarrow n=\frac{798}{21} \\ \Rightarrow n=38 \end{gathered}$$

Hence, the correct option is (B).