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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 ${S}_{n}=\frac{n}{2}\left(2a+\left(n-1\right)d\right)\phantom{\rule{0ex}{0ex}}{S}_{30}=\frac{30}{2}\left(2a+\left(30-1\right)d\right)\phantom{\rule{0ex}{0ex}}=\frac{30}{2}\left(2\left(1\right)+29\left(1\right)\right)\phantom{\rule{0ex}{0ex}}=15\left(2+29\right)\phantom{\rule{0ex}{0ex}}=15×31\phantom{\rule{0ex}{0ex}}=465$ Hence, the correct option is (B). (4) It is given that, t7 = 4 d = –4 n = 101 Now, ${t}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{t}_{7}=a+\left(7-1\right)d\phantom{\rule{0ex}{0ex}}⇒4=a+6\left(-4\right)\phantom{\rule{0ex}{0ex}}⇒4=a-24\phantom{\rule{0ex}{0ex}}⇒a=4+24\phantom{\rule{0ex}{0ex}}⇒a=28$ Hence, the correct option is (D). (5) It is given that, a = 3.5 d = 0 n = 101 ${t}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}=3.5+\left(n-1\right)\left(0\right)\phantom{\rule{0ex}{0ex}}=3.5$ Hence, the correct option is (B). (6) It is given that, a = –3 a2 = 4 We know that, ${a}_{2}=a+\left(2-1\right)d\phantom{\rule{0ex}{0ex}}⇒4=-3+d\phantom{\rule{0ex}{0ex}}⇒d=7$ Now, ${a}_{21}=a+\left(21-1\right)d\phantom{\rule{0ex}{0ex}}=-3+20\left(7\right)\phantom{\rule{0ex}{0ex}}=-3+140\phantom{\rule{0ex}{0ex}}=137$ Hence, the correct option is (C). (7) It is given that, d = 5 Now, ${t}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{t}_{18}-{t}_{13}=\left(a+\left(18-1\right)d\right)-\left(a+\left(13-1\right)d\right)\phantom{\rule{0ex}{0ex}}=\left(a+17d\right)-\left(a+12d\right)\phantom{\rule{0ex}{0ex}}=5d\phantom{\rule{0ex}{0ex}}=5\left(5\right)\phantom{\rule{0ex}{0ex}}=25$ Hence, the correct option is (C). (8) The given sequence is 3, 6, 9,... Here, a = 3 d = 3 n = 5 We know that, ${S}_{n}=\frac{n}{2}\left(2a+\left(n-1\right)d\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{S}_{5}=\frac{5}{2}\left(2a+\left(5-1\right)d\right)\phantom{\rule{0ex}{0ex}}=\frac{5}{2}\left(2\left(3\right)+4\left(3\right)\right)\phantom{\rule{0ex}{0ex}}=\frac{5}{2}\left(6+12\right)\phantom{\rule{0ex}{0ex}}=\frac{5}{2}\left(18\right)\phantom{\rule{0ex}{0ex}}=5\left(9\right)\phantom{\rule{0ex}{0ex}}=45$ Hence, the correct option is (A). (9) The given sequence is 15, 10, 5,... Here, a = 15 d = –5 We know that, ${S}_{n}=\frac{n}{2}\left(2a+\left(n-1\right)d\right)\phantom{\rule{0ex}{0ex}}{S}_{10}=\frac{10}{2}\left(2a+\left(10-1\right)d\right)\phantom{\rule{0ex}{0ex}}=5\left(2\left(15\right)+9\left(-5\right)\right)\phantom{\rule{0ex}{0ex}}=5\left(30-45\right)\phantom{\rule{0ex}{0ex}}=5\left(-15\right)\phantom{\rule{0ex}{0ex}}=-75$ 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, ${t}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}{S}_{n}=\frac{n}{2}\left(2a+\left(n-1\right)d\right)\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\frac{n}{2}\left(a+\left(a+\left(n-1\right)d\right)\right)\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\frac{n}{2}\left(a+{t}_{n}\right)\phantom{\rule{0ex}{0ex}}⇒399=\frac{n}{2}\left(1+20\right)\phantom{\rule{0ex}{0ex}}⇒399=\frac{21n}{2}\phantom{\rule{0ex}{0ex}}⇒21n=399×2\phantom{\rule{0ex}{0ex}}⇒n=\frac{798}{21}\phantom{\rule{0ex}{0ex}}⇒n=38\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ Hence, the correct option is (B).

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