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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 tn = ....
(A) 0 (B) 3.5 (C) 103.5 (D) 14.5

(6) In an A.P. first two terms are –3, 4 then 21st 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. 1st 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

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Solution

(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

Sn=n22a+n-1dS30=3022a+30-1d =30221+291 =152+29 =15×31 =465

Hence, the correct option is (B).

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

Now,
tn=a+n-1dt7=a+7-1d4=a+6-44=a-24a=4+24a=28

Hence, the correct option is (D).

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

tn=a+n-1d =3.5+n-10 =3.5

Hence, the correct option is (B).

(6) It is given that,
a = –3
a2 = 4

We know that,
a2=a+2-1d4=-3+dd=7

Now,
a21=a+21-1d =-3+207 =-3+140 =137

Hence, the correct option is (C).

(7) It is given that,
d = 5

Now,
tn=a+n-1dt18-t13=a+18-1d-a+13-1d =a+17d-a+12d =5d =55 =25

Hence, the correct option is (C).

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

We know that,
Sn=n22a+n-1dS5=522a+5-1d =5223+43 =526+12 =5218 =59 =45

Hence, the correct option is (A).

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

We know that,
Sn=n22a+n-1dS10=1022a+10-1d =5215+9-5 =530-45 =5-15 =-75

Hence, the correct option is (A).

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

We know that,
tn=a+n-1dSn=n22a+n-1dSn=n2a+a+n-1dSn=n2a+tn399=n21+20399=21n221n=399×2n=79821n=38

Hence, the correct option is (B).

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