Arithmetic Progressions
Section A: Multiple Choice Questions
1. The nth term of an A.P. is given by \(a_n = 3 + 4n\). The common difference is:
Solution Common difference \(d = a_2 - a_1\).
\(a_1 = 3 + 4(1) = 7\).
\(a_2 = 3 + 4(2) = 11\).
\(d = 11 - 7 = 4\).
\(a_1 = 3 + 4(1) = 7\).
\(a_2 = 3 + 4(2) = 11\).
\(d = 11 - 7 = 4\).
2. If p, q, r and s are in A.P. then \(r - q\) is:
Solution In an A.P., the difference between consecutive terms is constant.
\(q - p = r - q = s - r = d\).
Therefore, \(r - q = s - r\).
\(q - p = r - q = s - r = d\).
Therefore, \(r - q = s - r\).
3. If the sum of three numbers in an A.P. is 9 and their product is 24, then the numbers are:
Solution Let terms be \(a-d, a, a+d\).
Sum: \(3a = 9 \Rightarrow a = 3\).
Product: \((3-d)(3)(3+d) = 24 \Rightarrow 3(9-d^2) = 24 \Rightarrow 9-d^2 = 8 \Rightarrow d^2 = 1 \Rightarrow d = \pm 1\).
Terms are 2, 3, 4.
Sum: \(3a = 9 \Rightarrow a = 3\).
Product: \((3-d)(3)(3+d) = 24 \Rightarrow 3(9-d^2) = 24 \Rightarrow 9-d^2 = 8 \Rightarrow d^2 = 1 \Rightarrow d = \pm 1\).
Terms are 2, 3, 4.
4. The \((n-1)^{th}\) term of an A.P. given by 7, 12, 17, 22, ... is:
Solution \(a = 7, d = 5\).
\(a_{n-1} = a + [(n-1)-1]d = 7 + (n-2)5 = 7 + 5n - 10 = 5n - 3\).
\(a_{n-1} = a + [(n-1)-1]d = 7 + (n-2)5 = 7 + 5n - 10 = 5n - 3\).
5. The \(10^{th}\) term from the end of the A.P. -5, -10, -15, ..., -1000 is:
Solution From the end, first term \(l = -1000\), common difference \(d = -(-5) = 5\).
But from reverse, new 'd' is +5. Wait, original series: -5, -10, -15... d = -5.
Reverse series: -1000, -995... d = +5.
\(a_{10} = l - (n-1)d = -1000 - (9)(-5) = -1000 + 45 = -955\). (Note: Formula used in image is \(l - (n-1)d\) where d is original diff).
But from reverse, new 'd' is +5. Wait, original series: -5, -10, -15... d = -5.
Reverse series: -1000, -995... d = +5.
\(a_{10} = l - (n-1)d = -1000 - (9)(-5) = -1000 + 45 = -955\). (Note: Formula used in image is \(l - (n-1)d\) where d is original diff).
6. Find the sum of 12 terms of an A.P. whose nth term is given by \(a_n = 3n + 4\).
Solution \(a_1 = 3(1) + 4 = 7\). \(a_2 = 3(2) + 4 = 10\). \(d = 3\).
\(S_{12} = \frac{12}{2}[2(7) + (11)(3)] = 6[14 + 33] = 6 \times 47 = 282\).
\(S_{12} = \frac{12}{2}[2(7) + (11)(3)] = 6[14 + 33] = 6 \times 47 = 282\).
7. The sum of all two-digit odd numbers is:
Solution Series: 11, 13, 15, ..., 99.
\(a = 11, l = 99, d = 2\).
\(99 = 11 + (n-1)2 \Rightarrow 88 = 2(n-1) \Rightarrow n-1 = 44 \Rightarrow n = 45\).
\(S_{45} = \frac{45}{2}(11 + 99) = \frac{45}{2}(110) = 45 \times 55 = 2475\).
\(a = 11, l = 99, d = 2\).
\(99 = 11 + (n-1)2 \Rightarrow 88 = 2(n-1) \Rightarrow n-1 = 44 \Rightarrow n = 45\).
\(S_{45} = \frac{45}{2}(11 + 99) = \frac{45}{2}(110) = 45 \times 55 = 2475\).
8. If \((p+q)^{th}\) term of an A.P. is m and \((p-q)^{th}\) term is n, then pth term is:
Solution \(a_{p+q} = m\), \(a_{p-q} = n\).
\(a + (p+q-1)d = m\) ...(i)
\(a + (p-q-1)d = n\) ...(ii)
Adding (i) and (ii): \(2a + (2p-2)d = m+n \Rightarrow 2[a + (p-1)d] = m+n\).
\(a_p = \frac{m+n}{2}\).
\(a + (p+q-1)d = m\) ...(i)
\(a + (p-q-1)d = n\) ...(ii)
Adding (i) and (ii): \(2a + (2p-2)d = m+n \Rightarrow 2[a + (p-1)d] = m+n\).
\(a_p = \frac{m+n}{2}\).
9. The number of multiples lie between n and \(n^2\) which are divisible by n is:
Solution Multiples of n from 1 to \(n^2\) are \(n, 2n, 3n, ..., n^2\). Total n numbers.
Between n and \(n^2\) excludes the first and last terms.
Total = \(n - 2\).
Between n and \(n^2\) excludes the first and last terms.
Total = \(n - 2\).
10. If a, b, c, d, e are in A.P., then the value of \(a - 4b + 6c - 4d + e\) is:
Solution Let terms be \(a, a+x, a+2x, a+3x, a+4x\).
\(a - 4(a+x) + 6(a+2x) - 4(a+3x) + (a+4x)\).
Coefficients of 'a': \(1 - 4 + 6 - 4 + 1 = 0\).
Coefficients of 'x': \(-4 + 12 - 12 + 4 = 0\).
Result is 0.
\(a - 4(a+x) + 6(a+2x) - 4(a+3x) + (a+4x)\).
Coefficients of 'a': \(1 - 4 + 6 - 4 + 1 = 0\).
Coefficients of 'x': \(-4 + 12 - 12 + 4 = 0\).
Result is 0.
Section B: Assertion-Reason Questions
Directions:
(a) Both A and R are true and R is correct explanation of A.
(b) Both A and R are true but R is not correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
(a) Both A and R are true and R is correct explanation of A.
(b) Both A and R are true but R is not correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
11. Assertion (A): Let the positive numbers a, b, c be in A.P., then \(\frac{1}{bc}, \frac{1}{ac}, \frac{1}{ab}\) are also in A.P.
Reason (R): If each term of an A.P. is divided by abc, then the resulting sequence is also in A.P.
Reason (R): If each term of an A.P. is divided by abc, then the resulting sequence is also in A.P.
Solution Dividing the AP terms a, b, c by abc gives \(\frac{1}{bc}, \frac{1}{ac}, \frac{1}{ab}\). This operation maintains AP property. Thus A and R are true, and R explains A.
12. Assertion (A): If \(S_n\) is the sum of the first n terms of an AP, then its nth term \(a_n\) is given by \(a_n = S_n - S_{n-1}\).
Reason (R): The 10th term of the A.P. 5, 8, 11, 14, ... is 35.
Reason (R): The 10th term of the A.P. 5, 8, 11, 14, ... is 35.
Solution Assertion is a standard formula (True).
Reason: \(a_{10} = 5 + 9(3) = 32\). Reason says 35, which is False.
Reason: \(a_{10} = 5 + 9(3) = 32\). Reason says 35, which is False.
13. Assertion (A): The sum of the series with the nth term, \(a_n = (9 - 5n)\) is (465), when n = 15.
Reason (R): Given series is in A.P. and sum of n terms of an A.P. is \(S_n = \frac{n}{2}[2a + (n-1)d]\).
Reason (R): Given series is in A.P. and sum of n terms of an A.P. is \(S_n = \frac{n}{2}[2a + (n-1)d]\).
Solution Reason is True (formula).
Assertion: \(S_{15} = \frac{15}{2}(a_1 + a_{15})\). \(a_1 = 4, a_{15} = 9-75 = -66\). Sum = \(\frac{15}{2}(4-66) = \frac{15}{2}(-62) = -465\). Assertion says 465 (positive), so it is False.
Assertion: \(S_{15} = \frac{15}{2}(a_1 + a_{15})\). \(a_1 = 4, a_{15} = 9-75 = -66\). Sum = \(\frac{15}{2}(4-66) = \frac{15}{2}(-62) = -465\). Assertion says 465 (positive), so it is False.
14. Assertion (A): The value of n is 18, if \(a = 10, d = 5, a_n = 95\).
Reason (R): The formula of general term is \(a_n = a + (n-1)d\).
Reason (R): The formula of general term is \(a_n = a + (n-1)d\).
Solution \(95 = 10 + (n-1)5 \Rightarrow 85 = 5(n-1) \Rightarrow 17 = n-1 \Rightarrow n = 18\). Assertion is True. Reason is the formula used.
15. Assertion (A): The 11th term of an AP 7, 9, 11, 13, .......is 67.
Reason (R): if \(S_n\) is the sum of first n terms of an AP then its nth term \(a_n\) is given by \(a_n = S_n - S_{n-1}\).
Reason (R): if \(S_n\) is the sum of first n terms of an AP then its nth term \(a_n\) is given by \(a_n = S_n - S_{n-1}\).
Solution Reason is True.
Assertion: \(a_{11} = 7 + 10(2) = 27\). Assertion says 67, which is False.
Assertion: \(a_{11} = 7 + 10(2) = 27\). Assertion says 67, which is False.
16. Assertion (A): The first term of an AP is m and the common difference is p then the \(13^{th}\) term is a+10p.
Reason (R): In an AP, \(a_n = S_n - S_{n-1}\).
Reason (R): In an AP, \(a_n = S_n - S_{n-1}\).
Solution Reason is True.
Assertion: \(13^{th}\) term should be \(m + 12p\). Assertion says \(a+10p\), which is False.
Assertion: \(13^{th}\) term should be \(m + 12p\). Assertion says \(a+10p\), which is False.
17. Assertion (A): In an AP, \(S_n = n^2 + n\) then \(a_{20} = 40\).
Reason (R): In an AP \(d = a_n - a_{n-1}\).
Reason (R): In an AP \(d = a_n - a_{n-1}\).
Solution Assertion: \(a_n = S_n - S_{n-1} = (n^2+n) - ((n-1)^2+(n-1)) = 2n\).
So \(a_{20} = 2(20) = 40\). Assertion is True.
Reason is True definition of common difference. But R does not explain calculation of \(a_{20}\) from \(S_n\).
So \(a_{20} = 2(20) = 40\). Assertion is True.
Reason is True definition of common difference. But R does not explain calculation of \(a_{20}\) from \(S_n\).
18. Assertion (A): If the nth term of an AP is \(2n^2 - 1\), then the sum of the first n terms is \(n^3\).
Reason (R): If a, l and n are first, last and number of terms of an AP respectively, then \(S_n = \frac{n}{2}(a + l)\).
Reason (R): If a, l and n are first, last and number of terms of an AP respectively, then \(S_n = \frac{n}{2}(a + l)\).
Solution Reason is True.
Assertion: If \(a_n\) involves \(n^2\), it's not a standard AP (linear nth term). Sum of \(2n^2-1\) is not \(n^3\). Assertion is False.
Assertion: If \(a_n\) involves \(n^2\), it's not a standard AP (linear nth term). Sum of \(2n^2-1\) is not \(n^3\). Assertion is False.
19. Assertion (A): The value of n, if \(a = 10, d = 5, a_n = 95\).
Reason (R): The formula of general term is \(a_n = a + (n-1)d\).
Reason (R): The formula of general term is \(a_n = a + (n-1)d\).
Solution (Duplicate of Q14). \(95 = 10 + (n-1)5 \Rightarrow n=18\). Assertion implies n is found correctly (though not explicitly stated in text "is 18"). Assuming it refers to finding n=18. A is True, R is True and explains it.
20. Assertion (A): The nth term of an arithmetic progression is given by \(a_n = a + (n-1)d\).
Reason (R): In an AP, the difference between any two consecutive terms is constant.
Reason (R): In an AP, the difference between any two consecutive terms is constant.
Solution Both are fundamental definitions. The constant difference (Reason) is the basis for deriving the nth term formula (Assertion). Thus, R explains A.