

InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
251. |
1 + (0.04) + (0.04)2 + ………………. ∞ = A) 24/25B) 1 C) 0.04 D) 25/24 |
Answer» Correct option is (D) 25/24 \(\because1,0.04,(0.04)^2,.......\) will form a G.P. whose first term is a = 1 & common difference is r = 0.04 < 1 \(\therefore S_\infty=1+0.04+(0.04)^2+……\) \(=\frac1{1-0.04}\) \((\because S_\infty=\frac a{1-r},r<1)\) \(=\frac1{0.96}\) \(=\frac{100}{96}\) \(=\frac{25}{24}\) Hence, \(1+0.04+(0.04)^2+……\) \(=\frac{25}{24}\) Correct option is D) 25/24 |
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252. |
If |x| < 1,y = x – x2 + x3 – x4 + ……………… ∞,then the value of x in terms of y isA) 1 - y/yB) y/1 - yC) y/y + 1D) y/y + 1 |
Answer» Correct option is (B) y/1 - y Since, \(x,-x^2,x^3,-x^4,......\) will form a G.P. whose common difference is -x & first term is x. Now, \(y=x-x^2+x^3-x^4+......\) \(upto\,\infty\,\text{terms (Given)}\) \(=\frac x{1-(-x)}\) \((\because|x|<1\Rightarrow S_\infty=\frac a{1-r}\) here r = -x & a = x) \(=\frac x{1+x}\) \(\Rightarrow\frac1y=\frac{1+x}x\) \(\Rightarrow\frac1y=\frac{1}x+1\) \(\Rightarrow\frac1x=\frac1y-1=\frac{1-y}y\) \(\Rightarrow\) \(x=\frac{y}{1-y}\) Correct option is B) y/1 - y |
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253. |
Coefficient of x2 in (x + 1) (x + 2) (x + 3) (x + 4) (x + 5) is ……………….. A) 215 B) 205 C) 200D) 225 |
Answer» Correct option is (D) 225 (x+1) (x+2) (x+3) (x+4) (x+5) \(=(x^2+3x+2)(x^2+7x+12)(x+5)\) \(\therefore\) Coefficient of \(x^2\) in \((x+1)(x+2)(x+3)(x+4)(x+5)\) \(=12.5+2.5+21.5+14.1+36.1\) \(=60+10+105+14+36\) \(=120+105=225\) Correct option is D) 225 |
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254. |
1/2 + 1/4 + 1/8 + 1/16 + ............= x, 1/3 + 1/9 + 1/21 + 1/81 + y , thenA) x2 = yB) 2x + 4y = 4 C) x = y D) x + y = 0 |
Answer» Correct option is B) 2x + 4y = 4 |
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255. |
The sum to infinity of 1/7 + 2/72 + 1/73 + 2/74 +……………. isA) 5/48B) 7/24C) 3/16D) 1/5 |
Answer» Correct option is (C) 3/16 \(\because\frac17,\frac1{7^3},........\) will form a G.P. of common ratio \(=\frac1{7^2}<1\) And \(\frac2{7^2},\frac2{7^4},........\) will form a G.P. of common ratio \(=\frac1{7^2}<1\) We know that if r < 1 then sum of infinity terms in a G.P. is given by \(S_\infty=\frac a{1-r}.\) Now, \(\frac{1}{7}+\frac{2}{7^2}+\frac{1}{7^3}+\frac{2}{7^4}+ ........\) \(=(\frac17+\frac1{7^3}+.......)+(\frac2{7^2}+\frac2{7^4}+........)\) \(=\cfrac{\frac{1}{7}}{1-\frac{1}{7^2}}+\cfrac{\frac{2}{7^2}}{1-\frac{1}{7^2}}\) \(=\cfrac{\frac{1}{7}}{\frac{49-1}{49}}+\cfrac{\frac{2}{49}}{\frac{49-1}{49}}\) \(=\frac{1}{7}\times\frac{49}{48}+\frac{2}{49}\times\frac{49}{48}\) \(=\frac{7}{48}+\frac{2}{48}\) \(=\frac{9}{48}=\frac{3}{16}\) Hence, \(\frac{1}{7}+\frac{2}{7^2}+\frac{1}{7^3}+\frac{2}{7^4}+.......\) \(=\frac{3}{16}\) Correct option is C) 3/16 |
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256. |
The least value of n so that 1 + 32 + 33 + 3 + ………….. n terms is greater than 2000 is ……………. A) 8B) 7 C) 9 D) 6 |
Answer» Correct option is (A) 8 \(\because\) \(1+3+3^2+3^3+.....\) will form a G.P. with common ratio 3 whose sum of n terms is given by \(\frac{a(r^n-1)}{r-1}.\) \(S_n=1+3+3^2+3^3+......+n\) terms \(=\frac{1(3^n-1)}{3-1}=\frac{3^n-1}2\) Let \(S_n>2000\) \(\Rightarrow\) \(\frac{3^n-1}2>2000\) \(\Rightarrow3^n-1>4000\) \(\Rightarrow3^n>4001\) \(\because3^7=2187<4001\) But \(3^8=6561>4001\) Hence, least value of n is 8 so that the sum \(1+3+3^2+3^3+.....\) is greater than 2000. Correct option is A) 8 |
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257. |
Sum to n terms of 1.2.3 + 2.3.4 + 3.4.5 + ……………… is .A) \(\cfrac{n(n+1)(n+2)(n+3)}{6}\) B) \(\cfrac{n(n+1)(n+2)}{2}\)C) \(\cfrac{n(n+1)(n+2)(n+3)}{4}\)D) \(\cfrac{n(n+1)(n+2)(n+3)}{8}\) |
Answer» Correct option is (C) \(\frac{n(n+1)(n+2)(n+3)}{4}\) \(S_n=\) 1.2.3 + 2.3.4 + 3.4.5 + …….. + upto n terms = 1.2.3 + 2.3.4 + 3.4.5 + …….. + n (n+1) (n+2) \(=\sum n(n+1)(n+2)\) \(=\sum n(n^2+3n+2)\) \(=\sum n^3+3n^2+2n\) \(=\sum n^3+3\sum n^2+2\sum n\) \(=\left(\frac{n(n+1)}2\right)^2+3\left(\frac{n(n+1)(2n+1)}6\right)+\frac{2n(n+1)}2\) \(=n(n+1)\left(\frac{n(n+1)}4+\frac{2n+1}2+1\right)\) \(=\frac{n(n+1)}4(n^2+n+4n+2+4)\) \(=\frac{n(n+1)}4(n^2+5n+6)\) \(=\frac{n(n+1)(n+2)(n+3)}{4}\) Correct option is C) \(\cfrac{n(n+1)(n+2)(n+3)}{4}\) |
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258. |
Which term of 7 + 11 +15 + ………………. is 67? A) 20 B) 6 C) 16 D) 10 |
Answer» Correct option is (C) 16 7, 11, 15, ......... will form an A.P. whose first term is a = 7 & common difference is 4. Let \(n^{th}\) term of A.P. will be 67. \(\therefore a_n=67\) \(\Rightarrow a+(n-1)d=67\) \(\Rightarrow7+(n-1)4=67\) \((\because\) a = 7 & d = 4) \(\Rightarrow4(n-1)=67-7=60\) \(\Rightarrow n-1=\frac{60}4=15\) \(\Rightarrow n=15+1=16\) Hence, \(16^{th}\) term of 7, 11, 15, ......... will be 67. Correct option is C) 16 |
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259. |
The greatest value of n so that 1 + 5 + 52 + 53 + …………. n terms is less than 4321 is ……………….. A) 7 B) 6 C) 9 D) 8 |
Answer» Correct option is (B) 6 Let \(1 + 5 + 5^2 + 5^3 +.......+n\) terms < 4321 \(\Rightarrow\frac{1(5^n-1)}{5-1}<4321\) \(\Rightarrow5^n-1<4321\times4\) \(\Rightarrow5^n-1<17284\) \(\Rightarrow5^n<17285\) \(\because5^6=15625<17285\) But \(5^7=78125>17285\) Hence, greatest value of n is 6 so that the sum \(1+5+5^2+5^3+.......\) is less than 4321. Correct option is B) 6 |
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260. |
nth 5/1.2 + 7/2.3 x 1/3 + 9/3.4 x 1/32 + 11/4.5 x 1/33 term of the seriesA) \(\cfrac{2n+4}{n(n+1)}\) x \(\cfrac{1}{3^n}\)B) \(\cfrac{2n+3}{n(n+1)}\) x \(\cfrac{1}{3^{n+1}}\)C) \(\cfrac{2n+3}{n(n+1)}\) x \(\cfrac{1}{3^n}\)D) \(\cfrac{2n+3}{n(n+1)}\) x \(\cfrac{1}{3^{n-1}}\) |
Answer» Correct option is D) \(\cfrac{2n+3}{n(n+1)}\) x \(\cfrac{1}{3^{n-1}}\) |
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261. |
Which series is equal to \(\cfrac{n(n+1)}2\) ?A) 13 + 23 + 33 + + n3 B) 1.2 + 2.3 + 3.4 + + n(n + 1) C) 1 + 2 + 3+ ……+ n D) 12 + 22 + 32 + ……………… + n2 |
Answer» Correct option is (C) 1 + 2 + 3+ ……+ n \(\sum n=1+2+3+.......+n\) \(=\frac{n(n+1)}{2}\) Correct option is C) 1 + 2 + 3+ ……+ n |
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262. |
Which term of 0.0004,0.02,0.1,……………… is 12 ? A) 7 B) 5 C) 8 D) 6 |
Answer» Correct option is (D) 6 Let \(a_1=0.004=\frac4{1000}\) \(a_2=0.02=\frac2{100}\) \(a_3=0.1=\frac1{10}\) Now, \(\frac{a_2}{a_1}=\cfrac{\frac{2}{100}}{\frac{4}{1000}}\) \(=\frac{2}{100}\times\frac{1000}{4}\) \(=\frac{10}{2}=5\) \(\frac{a_3}{a_2}=\cfrac{\frac{1}{10}}{\frac{2}{100}}\) \(=\frac{1}{10}\times\frac{100}{2}\) \(=\frac{10}{2}=5\) Hence, 0.004, 0.02, 0.1, ......… is a geometric progression whose first term is \(a=a_1=0.004=\frac4{1000}\) & common ratio \(=r=\frac{a_2}{a_1}=5\) Let \(n^{th}\) term of the G.P. is 12.5. i.e., \(a_n=12.5\) \(\Rightarrow ar^{n-1}=12.5\) \(\Rightarrow\frac4{1000}\times5^{n-1}=\frac{125}{10}\) \(\Rightarrow5^{n-1}=\frac{125}{10}\times\frac{1000}{4}\) \(=125\times25\) \(=5^3\times5^2=5^5\) \(\therefore n-1=5\) \(\Rightarrow n=5+1=6\) Hence, \(6^{th}\) term of given sequence is 6. Correct option is D) 6 |
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263. |
How many terms are there in the A.P. whose first and fifth terms are -14 and 2 respectively and the sum of the terms is 40? |
Answer» Given, the first term (a) = -14, Filth term (a5) = 2, Sum of terms (Sn) = 40 of the A.P. If the common difference is taken as d. Then, a5 = a + 4d ⟹ 2 = -14 + 4d ⟹ 2 + 14 = 4d ⟹ 4d = 16 ⟹ d = 4 Next, we know that Sn = \(\frac{n}{2}\)[2a + (n − 1)d] Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms Now, on substituting the values in Sn ⟹ 40 = \(\frac{n}{2}\)[2(−14) + (n − 1)(4)] ⟹ 40 = \(\frac{n}{2}\)[−28 + (4n − 4)] ⟹ 40 = \(\frac{n}{2}\)[−32 + 4n] ⟹ 40(2) = – 32n + 4n2 So, we get the following quadratic equation, 4n2 – 32n – 80 = 0 ⟹ n2 – 8n + 20 = 0 On solving by factorization method, we get 4n2 – 10n + 2n + 20 = 0 ⟹ n(n – 10) + 2( n – 10 ) = 0 ⟹ (n + 2)(n – 10) = 0 Either, n + 2 = 0 ⟹ n = -2 Or, n – 10 = 0 ⟹ n = 10 Since the number of terms cannot be negative. Therefore, the number of terms (n) is 10. |
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264. |
How many terms of the sequence 18, 16, 14…. should be taken so that their sum is zero. |
Answer» Given AP. is 18, 16, 14, … We know that, Sn = \(\frac{n}{2}\)[2a + (n − 1)d] Here, The first term (a) = 18 The sum of n terms (Sn) = 0 (given) Common difference of the A.P. (d) = a2 – a1 = 16 – 18 = – 2 So, on substituting the values in Sn ⟹ 0 = \(\frac{n}{2}\)[2(18) + (n − 1)(−2)] ⟹ 0 = \(\frac{n}{2}\)[36 + (−2n + 2)] ⟹ 0 = \(\frac{n}{2}\)[38 − 2n] Further, \(\frac{n}{2}\) ⟹ n = 0 Or, 38 – 2n = 0 ⟹ 2n = 38 ⟹ n = 19 Since, the number of terms cannot be zero, hence the number of terms (n) should be 19. |
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265. |
If the nth term of the A.P. 9, 7, 5, ...... is same as the nth term of the A.P. 15, 12, 9, ...... find n. |
Answer» A = 9, D = 7- 9 = - 2 a = 15, d = - 3 An = an A+(n - 1)D = a +(n -1)d 9 – (n -1)2 = 15 – (n -1)3 (n -1)(3 - 2) = 6 n - 1 = 6 n = 7 |
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266. |
If (m+1)th term of an A.P. is twice the (n+1)th term, prove that (3m+1)th term is twice (m+n+1)th term. |
Answer» Given: (m+1)th term of an A.P. is twice the (n+1)th term a(m +1) = 2a(n +1) To Prove: (3m+1)th term is twice (m+n+1)th term a(3m + 1) = 2a(m +n +1) Proof: a(m +1) = 2a(n +1) ⇒ a + (m + 1 – 1) d = 2a + 2(n +1 -1)d ⇒ - a = 2nd – md ⇒ a = md - 2nd (i) LHS: a3m+1 = a + (3m + 1 -1)d = md – 2nd + 3md = 2d(2m - n) RHS: 2a(m + n + 1) = 2[a +(m +n +1 -1)d] = 2[md - 2nd + md + nd] = 2d(2m - n) LHS = RHS Hence, proved |
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267. |
The 24th term of an A.P. is twice its 10th term. Show that its 72nd term is 4 times its 15th term. |
Answer» Given: a24 = 2a10 a +23d = 2(a +9d) 5d = a (i) To Prove: a72 = 4a15 Proof: L.H.S. = a72 = a +71d = 5d +71d [from (i)] = 76d R.H.S = 4a15 = 4(a +14d) = 4a +56d = 4(5d) +56d [from (i)] = 20d +56d = 76d Since, L.H.S = R.H.S. Hence proved |
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268. |
In a certain A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term. |
Answer» Given: a24 = 2a10 To Prove: a72 = 2a34 Proof: a24 = 2a10 a + (24 –1)d = 2a + 2(10 –1)d 23d – 18d = a a = 5d LHS= a72 = a + 71d = 5d +71d = 76d RHS = a34 = 2a + 2(33)d = 10d + 66d = 76d Since, LHS = RHS Hence, proved |
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269. |
Which term of the A.P. 3, 10, 17, 24, ... will be 84 more than its 13th term ? |
Answer» Given: 3, 10, 17, 24, ... First we need to calculate 13th term. We know that an = a + (n – 1)d Here, a = 3, d = 10 – 3 = 7 and n = 13 So, a13 = 3 + (13 – 1)7 ⇒ a13 = 3 + 12 × 7 ⇒ a13 = 3 + 84 ⇒ a13 = 87 Now, the term is 84 more than a13 is 84 + 87 = 171 Now, a + (n – 1)d = 171 ⇒ 3 + (n – 1)7 = 171 ⇒ 3 + 7n – 7 = 171 ⇒ 7n = 171 + 7 – 3 ⇒ 7n = 175 ⇒ n = 25 Hence, the 25th term is 84 more than the 13th term. |
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270. |
The 4th term of an A.P. is zero. Prove that its 25th term is triple its 11th term. |
Answer» Given: a4 = 0 To Prove: a25 = 3 × a11 Now, a4 = 0 ⇒ a + 3d = 0 ⇒ a = –3d We know that, an = a + (n – 1)d a11 = –3d + (11 – 1)d [from (i)] a11 = –3d + 10d a11 = 7d …(ii) Now, a25 = a + (25 – 1)d a25 = –3d + 24d [from(i)] a25 = 21d a25 = 3 × 7d a25 = 3 × a11 [from(ii)] Hence Proved |
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271. |
The nth term of an AP is (7 – 4n). Find its common difference. |
Answer» Nth term = an = 7 – 4n a(n-1) = 7 – 4(n – 1) = 11 – 4n Common difference = d = an – a(n-1) = (7 – 4n) – (11 – 4n) = 7 – 4n – 11 + 4n = -4 Therefore, common difference is -4. |
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272. |
The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 34. Find the first term and the common difference of the A.P.? |
Answer» Given: a4 + a8 = 24 …(i) a6 + a10 = 34 …(ii) a4 = a + 3d a8 = a + 7d a6 = a + 5d a10 = a + 9d Put the value of a4 and a8 in (i) (a + 3d) + (a + 7d) = 24 a + 5d = 12 …(iii) Put the value of a6 and a10 in (ii) (a +5d) + (a +9d) = 34 a +7d = 17 …(iv) Subtracting (iii) from (iv), we get 2d = 5 d = \(\frac{5}{2}\) Putting the value of d in (iii), we get a = 12 - \(\frac{25}{2}\) = \(\frac{-1}{2}\) Hence, first term is \(\frac{-1}{2}\) and common difference is \(\frac{5}{2}\) |
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273. |
The first term of an A.P. is 5 and its 100th term is - 292. Find the 50th term of this A.P. |
Answer» Given, a = 5 a100 = - 292 5 – 99d = - 292 d = \(\frac{-297}{-99}\) = 3 Now, 50th term, a50 = a + (50 – 1) d = 5 + (49) 3 = 5 + 147 = 152 Hence, 50th term of given A.P. is 152 |
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274. |
The first term of an A.P. is 5 and its 100th term is -292. Find the 50th term of this A.P. |
Answer» Given, an A.P whose a = 5 and a100 = -292 We know that an = a + (n – 1)d a100 = 5 + 99d = -292 99d = -297 d = -3 Hence, the 50th term is a50 = a + 49d = 5 + 49(-3) = 5 – 147 = -142 |
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275. |
Find a30 – a20 for the A.P. -9, -14, -19, -24. |
Answer» We know that an = a + (n – 1)d So, a30 – a20 = (a + 29d) – (a + 19) =10d Given A.P. -9, -14, -19, -24 Here, a = -9 and d = -14 – (-9) = = -14 + 9 = -5 So, a30 – a20 = 10d = 10(-5) = -50 |
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276. |
Write the expression an – ak for the A.P. a, a+d, a+2d, …..Hence, find the common difference of the A.P. for which(i) 11th term is 5 and 13th term is 79.(ii) a10 – a5 = 200(iii) 20th term is 10 more than the 18th term. |
Answer» Given A.P. a, a+d, a+2d, ….. So, an = a + (n-1)d = a + nd –d And, ak = a + (k-1)d = a + kd – d an – ak = (a + nd – d) – (a + kd – d) = (n – k)d (i) Given 11th term is 5 and 13th term is 79, Here n = 13 and k = 11, a13 – a11 = (13 – 11)d = 2d ⇒ 79 – 5 = 2d d = 74/2 = 37 (ii) Given, a10 – a5 = 200 ⇒ (10 – 5)d = 200 5d = 200 d = 40 (iii) Given, 20th term is 10 more than the 18th term. ⇒ a20 – a18 = 10 (20 – 18)d = 10 2d = 10 d = 5 |
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277. |
Find a30 – a20 for the A.P. a, a+d, a+2d, a+3d, …… |
Answer» We know that an = a + (n – 1)d So, a30 – a20 = (a + 29d) – (a + 19) =10d Given A.P. a, a+d, a+2d, a+3d, …… So, a30 – a20 = (a + 29d) – (a + 19d = 10d |
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278. |
Find n if the given value of x is the nth term of the given A.P.-1, -3, -5, -7, …; x = -151 |
Answer» Given A.P. -1, -3, -5, -7, …., -151 Here, a = -1 d = -3 – (-1) = -2 Last term (nth term) = -151 We know that an = a + (n – 1)d ⇒ -151 = -1 + (n-1)(-2) -151 = -1 – 2n + 2 n = 152/2 n = 76 |
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279. |
If an A.P. consists of n terms with first term a and nth term ls how that the sum of the mth term from the beginning and the mth term from the end is (a+l). |
Answer» Given, a = a Last term, l = l We have to prove that the sum of the mth term from the beginning and mth term from the end is (a + 1) Now, mth term from the beginning, am(b) = a + (m – 1) l = a + ml – l (i) Again mth term from the end am(e) = l – (m – 1)l = l – ml + l = 2l – ml (ii) By adding (i) and (ii), we get a + ml – l + 2l – ml = a + l Hence, proved |
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280. |
Find n if the given value of x is the nth term of the given A.P.25, 50, 75, 100, ; x = 1000 |
Answer» Given A.P. 25, 50, 75, 100, …… ,1000 Here, a = 25 d = 50 – 25 = 25 Last term (nth term) = 1000 We know that an = a + (n – 1)d ⇒ 1000 = 25 + (n-1)25 1000 = 25 + 25n – 25 n = 1000/25 n = 40 |
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281. |
Find out which of the following sequences are arithmetic progressions. For those which are arithmetic progressions, find out the common difference.(i) 3, 3, 3, 3, ....(ii) p, p+90, p+180, p+270, .... where p = (999)999(iii) 1.0, 1.7, 2.4, 3.1, ......(iv) - 225, - 425, - 625, - 825, ....... |
Answer» (i) 3, 3, 3, 3, .... Common difference, d1= 3 – 3 = 0 Common difference, d2= 3 – 3 = 0 Since, d1 = d2 Therefore, it’s an A.P. with common difference, d = 0 (ii) p, p+90, p+180, p+270, .... where p = (999)999 Common difference, d1= p + 90 – p = 90 Common difference, d2 = p + 180 – p – 90 = 90 Since, d1 = d2 Therefore, it’s an A.P. with common difference, d = 90 (iii) 1.0, 1.7, 2.4, 3.1, ...... Common difference, d1= 1.7 – 1.0 = 0.7 Common difference, d2= 2.4 – 1.7 = 0.7 Since, d1 = d2 Therefore, it’s an A.P. with common difference, d = 0.7 (iv) - 225, - 425, - 625, - 825, ....... Common difference, d1 = - 425 + 225 = - 200 Common difference, d2 = - 625 + 425 = - 200 Since, d1 = d2 Therefore, it’s an A.P. with common difference, d = - 200 |
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282. |
The nth term of an A.P. is 6n+2. Find the common difference. |
Answer» an = 6n + 2 Put n = 1 a1 = 6(1) + 2 = 8 Put n = 2 a2 = 6(2) + 2 = 14 Common difference, d = a2 – a1 = 14 – 8 = 6 |
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283. |
Find out which of the following sequences are arithmetic progressions. For those which are arithmetic progressions, find out the common difference.(i) 10, 10 + 26, 10 + 27(ii) a + b, (a + b) + b, (a + b) + (b + 1), (a + 2) + (b + 1), (a + 2) + (b + 2), .....(iii) 12, 32, 52, 72, ......(iv) 12, 52, 72, 73, ....... |
Answer» (i) 10, 10 + 26, 10 + 27 Common difference, d1 = 10 + 26 – 10 = 26 = 64 Common difference, d2 = 10 + 27 – 10 – 26 = 26 (2 – 1) = 64 Since, d1 = d2 Therefore, it’s an A.P. with common difference, d = 64 (ii) a + b, (a + b) + b, (a + b) + (b + 1), (a + 2) + (b + 1), (a + 2) + (b + 2), ..... Common difference, d1 = (a + 1) + b – a – b = 1 Common difference, d2 = (a + 1) + (b + 1) – (a + 1) – b = 1 Since, d1 = d2 Therefore, it’s an A.P. with common difference, d = 1 (iii) 12, 32, 52, 72, ...... Common difference, d1 = 32 – 12 = 8 Common difference, d2 = 52 – 32 = 25 – 9 = 16 Since, d1 ≠ d2 Therefore, it’s not an A.P. (iv) 12, 52, 72, 73, ....... Common difference, d1 = 52 – 12 = 24 Common difference, d2 = 72 – 52 = 24 Since, d1 = d2 Therefore, it’s an A.P. with common difference, d = 24 |
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284. |
In an A.P., pth term is q and qth term is p, then its common difference is ………………A) 1 B) -1 C) 2 D) -2 |
Answer» Correct option is (B) -1 Given that \(p^{th}\) term in A.P. is q and \(q^{th}\) term in A.P. is p. i.e., in A.P. \(a_p=q\;\&\;a_q=p\) \(\Rightarrow\) a + (p - 1) d = q _____________(1) and a + (q - 1) d = p _____________(2) \((\because a_n=a+(n-1)d)\) Subtract equation (2) from (1), we obtain (a + (p - 1) d) - (a + (q - 1) d) = q - p \(\Rightarrow(p-1-(q-1))d=q-p\) \(\Rightarrow(p-q-1+1)d=q-p\) \(\Rightarrow(p-q)d=-(p-q)\) \(\Rightarrow d=\frac{-(p-q)}{p-q}=-1\) Hence, common difference of A.P. is -1. Correct option is B) -1 |
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285. |
If p, q, r are in A.P. and x, y, z in G.P., then Xq - r × Yr - p × Zp - q is equal to ………………….. A) p + q + r B) x × y × z C) 1 D) px + qy + rz |
Answer» Correct option is (C) 1 Given that p, q, r are in A.P. \(\therefore\) p+r = 2q _____________(1) Also given that x, y, z are in G.P. \(\therefore xz=y^2\) _____________(2) Now, \(x^{q-r}\times y^{r-p}\times z^{p-q}\) \(=x^{q-(2q-p)}\times y^{((2q-p)-p)}\times z^{p-q}\) (By putting r = 2q - p (from (1)) \(=x^{p-q}\times y^{2(q-p)}\times z^{p-q}\) \(=(xz)^{p-q}\times y^{2(q-p)}\) \((\because a^m.b^m=(ab)^m)\) \(=(y^2)^{p-q}\times y^{2(q-p)}\) (From (2)) \(=y^{2(p-q)}\times y^{-2(p-q)}\) \(=y^{2(p-q)-2(p-q)}\) \(=y^0=1\) Hence, \(x^{q-r}\times y^{r-p}\times z^{p-q}=1\) Correct option is C) 1 |
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286. |
Which term of the A.P. 21, 42, 63, 84, ....... is 420? |
Answer» a = 21, d = 21 An = a + (n – 1) d 420 = 21 + (n – 1) 21 420 = 21 + 21n – 21 420 = 21n n = 20 |
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287. |
Which term of the AP 21. 42, 63, 84, … is 420? |
Answer» Given A. P 21, 42, 63, 84, ……… a = 21, d = a2 – a1 = 42 – 21 = 21 We know that, nth term (an) = a +(n – 1)d And, given nth term = 420 21 + (n – 1)21 = 420 (n – 1)21 = 399 n – 1 = 399/21 = 19 n = 20 ∴ 20th term is 420. |
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288. |
Which term of the AP 4. 9, 14, …. is 254? |
Answer» Given A. P 4, 9, 14, ………… First term (a) = 4 Common difference (d) = a2 – a1 = 9 – 4 = 5 We know that, nth term (an) = a + (n – 1)d And, given nth term is 254 4 + (n – 1)5 = 254 (n – 1)∙5 = 250 n – 1 = 250/5 = 50 n = 51 ∴ 51th term in the A.P is 254. |
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289. |
Which term of the A.P. 4, 9, 14, ...... is 254? |
Answer» a = 4, d = 5 An = a + (n – 1) d 254 = 4 + (n – 1) 5 254 = 4 + 5n – 5 254 = -1 + 5n 255 = 5n n = 51 |
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290. |
Check whether 301 is a term of the list of numbers 5, 11, 17, 23, . . . |
Answer» We have : a2 – a1 = 11 – 5 = 6, a3 – a2 = 17 – 11 = 6, a4 – a3 = 23 – 17 = 6 As ak + 1 – ak is the same for k = 1, 2, 3, etc., the given list of numbers is an AP. Now, a = 5 and d = 6. Let 301 be a term, say, the nth term of the this AP. We know that an = a + (n – 1) d So, 301 = 5 + (n – 1) × 6 i.e., 301 = 6n – 1 So, n=302/6=151/3 But n should be a positive integer (Why?). So, 301 is not a term of the given list of numbers. |
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291. |
Determine the AP whose 3rd term is 5 and the 7th term is 9. |
Answer» We have a3 = a + (3 – 1) d = a + 2d = 5 .....(1) and a7 = a + (7 – 1) d = a + 6d = 9 ......(2) Solving the pair of linear equations (1) and (2), we get a = 3, d = 1 Hence, the required AP is 3, 4, 5, 6, 7, . . . |
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292. |
The nth term of a sequence is given by an = 2n + 7. Show that it is an A.P. Also, find its 7th term. |
Answer» Given, an = 2n + 7 We can find first five terms of this sequence by putting values of n from 1 to 5. When n = 1 : a1 = 2(1) + 7 ⇒ a1 = 2 + 7 ⇒ a1 = 9 When n = 2 : a2 = 2(2) + 7 ⇒ a2 = 4 + 7 ⇒ a2 = 11 When n = 3 : a3 = 2(3) + 7 ⇒ a3 = 6 + 7 ⇒ a3 = 13 When n = 4 : a4 = 2(4) + 7 ⇒ a4 = 8 + 7 ⇒ a4 = 15 When n = 5 : a5 = 2(5) + 7 ⇒ a5 = 10 + 7 ⇒ a5 = 17 ∴ First five terms of the sequence are 9, 11, 13, 15, 17. A.P is known for Arithmetic Progression whose common difference = an – an-1 Where n > 0 a1 = 9, a2 = 11, a3 = 13, a4 = 15, a5 = 17 Now, a2 – a1 = 11 – 9 = 2 a3 – a2 = 13 – 11 = 2 a4 – a3 = 15 – 13 = 2 a5 – a4 = 17 – 15 = 2 As, a2 – a1 = a3 – a2 = a4 – a3 = a5 – a4 The given sequence is A.P Common difference, d = a2 – a1 = 2 To find the seventh term of A.P, firstly find an We know, an = a + (n-1)d Where a is first term or a1 and d is common difference ∴ an = 3 + (n-1) 2 ⇒ an = 3 + 2n – 2 ⇒ an = 2n + 1 When n = 7 : a7 = 2(7) + 1 ⇒ a7 = 14 + 1 ⇒ a7 = 15 Hence, The 7th term of A.P. is 15 |
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293. |
Find four numbers in AP whose sum is 8 and the sum of whose squares is 216. |
Answer» (4, 6, 8, 10) or (10, 8, 6, 4) |
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294. |
What will be the value of x if 8x + 4 , 6x – 2 and 2x + 7 are in arithmetic progression?A) -15/2B) 15/2C) 15 D) -15 |
Answer» Correct option is (B) 15/2 Given that 8x+4, 6x – 2 and 2x+7 are in A.P. \(\therefore2(6x-2)=8x+4+2x+7\) \((\because\) If a, b & c are in A.P. then 2b = a+c) \(\Rightarrow12x-4=10x+11\) \(\Rightarrow12x-10x=11+4=15\) \(\Rightarrow2x=15\) \(\Rightarrow\) x = \(\frac{15}{2}\) Correct option is B) 15/2 |
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295. |
Write the expression for the common difference of an A.P. whose first term is a and nth term is b. |
Answer» The nth term of the A.P whose first term a1 and common difference is d is given by an = a + (n–1) d Here an is given as b so b = a + (n–1) d \(\cfrac{b - a}{n - 1}\) = d ⇒ d = \(\cfrac{b - a}{n - 1}\) |
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296. |
If the sum of n terms of an A.P. is Sn = 3n2 + 5n. Write its common difference. |
Answer» Let a1, a2, a3…….. an be the given A.P Given, sum of n terms = 3n2 + 5n Sn=3n2 + 5n……….(1) Putting n =1 in (1) Sn= 3x12 + 5x1 = 3 + 5 = 8 Sum of first 1 terms = first term first term = a = S1 = 8 Sn = 3n2 + 5n Putting n= 2 in….. (1) S2= 3 x 22 + 5 x 2 S2= 22 Sum of first two terms = first term + second term S2= a1 + a2 S2 – a1 = a2 a2= 22 – 8 = 14 Thus a1= 8, a2 = 14 d= common difference = 14 – 8= 6 |
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297. |
Write the first three terms of the APs when a and d are as given below:a = –5, d = –3 |
Answer» a = –5, d = –3 We know that, First three terms of AP are : a, a + d, a + 2d -5, – 5 + 1 (- 3), – 5 + 2 (- 3) – 5, – 8, – 11 |
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298. |
Write the first three terms of the APs when a and d are as given below:a = 2 , d = 1/√2 |
Answer» a = √2 , d = 1/√2 We know that, First three terms of AP are : a, a + d, a + 2d √2, √2+1/√2, √2+2/√2 √2, 3/√2, 4/√2 |
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299. |
Write the first three terms of the APs when a and d are as given below:a = 1/2, d = -1/6 |
Answer» a =1/2, d = -1/6 We know that, First three terms of AP are : a, a + d, a + 2d ½, ½ + (-1/6), ½ + 2 (-1/6) ½, 1/3, 1/6 |
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300. |
Verify that a, 2a + 1, 3a + 2, 4a + 3,…is an AP, and then write its next three terms. |
Answer» Here a1 = a a2 = 2a + 1 a3 = 3a + 2 a4 = 4a + 3 a2 – a1 = (2a + 1) – (a) = a + 1 a3 – a2 = (3a + 2) – (2a + 1) = a + 1 a4 – a3 = (4a + 3) – (3a+2) = a + 1 Since, difference of successive terms are equal, Hence, a, 2a + 1, 3a + 2, 4a + 3,… is an AP with common difference a+1. Therefore, the next three term will be, 4a + 3 +(a + 1), 4a + 3 + 2(a + 1), 4a + 3 + 3(a + 1) 5a + 4, 6a + 5, 7a + 6 |
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