CBSE Class 10 Maths HOTs Arithmetic Progressions Set 14

Multiple Choice Questions 

Question. If \( p, q, r \) are in AP, then \( p^3 + r^3 - 8q^3 \) is equal to
(a) 4pqr
(b) - 6pqr
(c) 2pqr
(d) 8pqr
Answer: (b) - 6pqr
Sol. \( \because p, q, r \) are in AP.
\( \therefore 2q = p + r \)
\( \implies p + r - 2q = 0 \)
\( \therefore p^3 + r^3 + (-2q)^3 = 3 \times p \times r \times (-2q) \)
[using if \( a + b + c = 0 \implies a^3 + b^3 + c^3 = 3abc \)]
\( \implies p^3 + r^3 - 8q^3 = - 6pqr \).

Question. If \( p - 1, p + 3, 3p - 1 \) are in AP, then p is equal to
(a) 2
(b) 3
(c) 4
(d) 5
Answer: (c) 4
Sol. \( \because p - 1, p + 3 \) and \( 3p - 1 \) are in AP.
\( \therefore 2(p + 3) = p - 1 + 3p - 1 \)
\( \implies 2p + 6 = 4p - 2 \)
\( \implies -2p = -8 \)
\( \implies p = 4 \).

Question. If the first term of an AP is 2 and the common difference is \( (-\frac{1}{2}) \), then its 12th term is
(a) \( 2 + 11(-\frac{1}{2}) \)
(b) \( 2 + 12(-\frac{1}{2}) \)
(c) \( 2 - 11(-\frac{1}{2}) \)
(d) \( 2 - 12(-\frac{1}{2}) \)
Answer: (a) \( 2 + 11(-\frac{1}{2}) \)
Sol. As, \( a_n = a + (n - 1)d \)
So, \( a_{12} = a + 11d \)
\( \implies a_{12} = 2 + 11 \times (-\frac{1}{2}) \)

Question. The 10th term of 10.0, 10.5, 11.0, 11.5 is
(a) 12
(b) 14.5
(c) 19
(d) 21.5
Answer: (b) 14.5
Sol. \( a = 10, d = 10.5 - 10 = 0.5 \)
\( a_{10} = a + 9d = 10 + 9 \times 0.5 = 14.5 \)

Question. In an AP, if the common difference (d) = - 4 and the seventh term (\( a_7 \)) is 4, then the first term is 
(a) 32
(b) 28
(c) 24
(d) 20
Answer: (b) 28
Sol. Let a = first term,
Given, d = - 4, \( a_7 = 4 \)
\( \implies a + (7 - 1)d = 4 \)
[\( \because \) nth term of an AP = \( a_n = a + (n - 1)d \)]
\( \implies a + 6d = 4 \)
\( \implies a + 6 \times (- 4) = 4 \)
\( \implies a - 24 = 4 \)
\( \implies a = 4 + 24 = 28 \)

Question. The nth term of the A.P. \( \frac{1}{m}, \frac{1 + m}{m}, \frac{1 + 2m}{m}, ... \) is
(a) \( \frac{1 + m(n - 1)}{m} \)
(b) \( \frac{1 - m(n - 1)}{m} \)
(c) \( \frac{1 + m(1 - n)}{m} \)
(d) \( \frac{1 + m(1 + n)}{n} \)
Answer: (a) \( \frac{1 + m(n - 1)}{m} \)
Sol. In series \( \frac{1}{m}, \frac{1 + m}{m}, \frac{1 + 2m}{m}, ... \)
\( a_1 = \frac{1}{m}, a_2 = \frac{1 + m}{m}, a_3 = \frac{1 + 2m}{m} \)
\( d = a_3 - a_2 = \frac{1 + 2m}{m} - \frac{1 + m}{m} = \frac{1 + 2m - 1 - m}{m} = \frac{m}{m} = 1 \)
nth term of the AP = \( a_1 + (n - 1)d \)
\( = \frac{1}{m} + (n - 1)1 \)
\( = \frac{1}{m} + n - 1 = \frac{1 + (n - 1)m}{m} \)

Question. For what value of p, are \( 2p + 1, 13, 5p - 3 \) three consecutive terms of an AP?
(a) 1
(b) 4
(c) 5
(d) 6
Answer: (b) 4
Sol. If terms are in AP, then \( 13 - (2p + 1) = (5p - 3) - 13 \)
\( \implies 13 - 2p - 1 = 5p - 3 - 13 \)
\( \implies 28 = 7p \)
\( \implies p = 4 \)

Question. What is the common difference of an A.P. in which \( a_{21} - a_7 = 84 \)? 
(a) 84
(b) 42
(c) 21
(d) 6
Answer: (d) 6
Sol. Let ‘d’ be the common difference of the AP whose first term is ‘a’
Now, \( a_{21} - a_7 = 84 \)
\( \implies (a + 20d) - (a + 6d) = 84 \)
\( \implies 20d - 6d = 84 \)
\( \implies 14d = 84 \)
\( \implies d = 6 \)

Very Short Answer Type Questions

Question. Find how many integers between 200 and 500 are divisible by 8.
Answer: Numbers divisible by 8 between 200 and 500 are 208, 216, 224, 232, ..., 496
Since, \( 216 - 208 = 224 - 216 = 8 \)
Therefore, the given integers are in A.P.
Let first term, a = 208,
common difference d = 224 – 216 = 8,
Number of terms, n = ?
Last term, \( a_n = 496 \)
We know that, \( a_n = a + (n – 1)d \)
\( \implies 496 = 208 + (n – 1)(8) \)
\( \implies 496 – 208 = (n – 1)(8) \)
\( \implies 288 = 8n – 8 \)
\( \implies 8n = 288 + 8 = 296 \)
\( \implies n = \frac{296}{8} = 37 \)
Therefore, there are 37 integers between 200 and 500 divisible by 8.

Question. Which term of the progression 65, 61, 57, 53, ... is the first negative term? 
Answer: 65, 61, 57, 53 ....
In this progression, first term is 65.
Also, second term – first term = 61 – 65 = –4
third term – second term = 57 – 61 = –4
Let nth term be the first negative term.
\( \therefore a_n = a + (n – 1)d \)
\( \implies a_n = 65 + (n – 1) \times (–4) \)
= 65 – 4(n – 1)
Now A.T.Q. \( a_n < 0 \)
\( \implies 65 – 4(n – 1) < 0 \)
\( \implies 65 – 4n + 4 < 0 \)
\( \implies 69 – 4n < 0 \)
\( \implies 4n > 69 \)
\( \implies n > \frac{69}{4} \)
\( \implies n > 17 \frac{1}{4} \)
\( \implies n \ge 18 \)
[\( \because \) n is a natural number]
Thus, 18th term of the given progression is the first negative term.

Question. The 6th term of an Arithmetic Progression (AP) is –10 and its 10th term is –26. Determine the 15th term of the AP.
Answer: Let Ist term of AP = a and common difference = d.
Now, \( a_6 = – 10 \)
\( \implies a + 5d = – 10 \) ...(i)
Also, \( a_{10} = – 26 \)
\( \implies a + 9d = – 26 \) ...(ii)
Subtract (i) from (ii), \( 4d = – 16 \implies d = – 4 \)
Substituting in (i), we get
\( a + 5 \times (– 4) = – 10 \implies a = 10 \)
Now, \( a_{15} = a + 14d = 10 + 14 \times (–4) = – 46 \)

Question. Is –150 a term of the AP 17, 12, 7, 2...?
Answer: 17, 12, 7, 2, ....
Here, a = 17, d = 12 – 17 = – 5
Let \( a_n = – 150 \)
\( a + (n – 1)d = – 150 \)
\( \implies 17 + (n – 1) (– 5) = – 150 \)
\( \implies (n – 1) (– 5) = – 150 – 17 \)
\( \implies (n – 1) (– 5) = – 167 \)
\( \implies n – 1 = \frac{167}{5} \)
\( \implies n = \frac{167}{5} + 1 \)
\( \implies n = \frac{167 + 5}{5} = \frac{172}{5} = 34 \frac{2}{5} \)
So, n is not a natural number.
\( \therefore \) – 150 is not a term of the given A.P.

Question. Find k, if the given value of x is the kth term of the given AP
(i) 25, 50, 75, 100, ..., x = 1000
(ii) \( 5 \frac{1}{2}, 11, 16 \frac{1}{2}, 22, ..., x = 550 \).
Answer: (i) a = 25, d = 50 – 25 = 25, x = 1000
A.T.Q., \( a_k = x \)
\( \implies a + (k – 1)d = 1000 \)
\( \implies 25 + (k – 1)25 = 1000 \)
\( \implies (k – 1)25 = 975 \)
\( \implies k – 1 = \frac{975}{25} \)
\( \implies k – 1 = 39 \)
\( \implies k = 40 \)
(ii) a = \( 5 \frac{1}{2} = \frac{11}{2} \), \( d = a_2 – a_1 = 11 – \frac{11}{2} = \frac{11}{2} \) and x = 550
A.T.Q., \( a_k = x \)
\( \implies a + (k – 1)d = 550 \)
\( \implies \frac{11}{2} + (k – 1) \frac{11}{2} = 550 \)
\( \implies \frac{11}{2} + \frac{11}{2}k – \frac{11}{2} = 550 \)
\( \implies \frac{11}{2}k = 550 \)
\( \implies k = \frac{550 \times 2}{11} = 100 \)

Question. Write the expression \( a_n – a_k \) for the AP: a, a + d, a + 2d, ... and find the common difference of the AP for which
(i) 11th term is 5 and 13th term is 79.
(ii) 20th term is 10 more than the 18th term.
Answer: \( a_n = a + (n – 1)d \); \( a_k = a + (k – 1)d \)
Now, \( a_n – a_k = [a + (n – 1)d] – [a + (k – 1)d] \)
= (n – 1)d – (k – 1)d
= (n – 1 – k + 1)d
\( \implies a_n – a_k = (n – k)d \) ...(i)
(i) Here \( a_{11} = 5, a_{13} = 79 \)
Taking n = 13, k = 11, eq. (i) becomes
\( a_{13} – a_{11} = (13 – 11)d \)
\( \implies 79 – 5 = 2d \implies d = 37 \)
(ii) Let \( a_{18} = x \).
\( \therefore a_{20} = x + 10 \)
Taking n = 20 and k = 18, equation (i) becomes
\( a_{20} – a_{18} = (20 – 18)d \)
\( \implies (x + 10) – x = 2d \implies d = 5 \)

Question. Determine k so that \( 4k + 8, 2k^2 + 3k + 6 \) and \( 3k^2 + 4k + 4 \) are three consecutive terms of an AP.
Answer: For consecutive terms of AP,
\( 2(2k^2 + 3k + 6) = (3k^2 + 4k + 4) + (4k + 8) \)
\( \implies 4k^2 + 6k + 12 = 3k^2 + 8k + 12 \)
\( \implies k^2 – 2k = 0 \)
\( \implies k(k – 2) = 0 \implies k = 0 \) or k = 2

Question. If 5 times the 5th term of an AP is equal to 10 times the 10th term, show that its 15th term is zero.
Answer: Let 1st term = a and common difference = d.
\( a_5 = a + 4d, a_{10} = a + 9d \)
According to the question, \( 5 \times a_5 = 10 \times a_{10} \)
\( \implies 5(a + 4d) = 10(a + 9d) \)
\( \implies 5a + 20d = 10a + 90d \)
\( \implies a = – 14d \)
Now \( a_{15} = a + 14d \)
\( \implies a_{15} = – 14d + 14d = 0 \).

Question. Find 10th term from end of the AP 4, 9, 14, .... , 254. [KVS]
Answer: 10th term from end of AP 4, 9, 14, ..., 254 is 10th term of the AP 254, 249, 244, ... 14, 9, 4
Here a = 254
d = 249 – 254 = –5
\( \therefore a_{10} = a + 9d \)
\( \implies a_{10} = 254 + 9 \times (–5) = 209 \)

Question. If x, y, z are in A.P., show that \( (x + 2y – z) (2y + z – x) (z + x – y) = 4xyz \).
Answer: \( \because \) x, y, z are in A.P.
\( \therefore 2y = x + z \)
Substituting \( 2y = x + z \)
L.H.S. is expressed as
(x + 2y – z) (2y + z – x) (z + x – y)
= (x + (x + z) – z) (x + z + z – x) (2y – y)
= 2x × 2z × y = 4xyz = R.H.S.

Question. If \( \frac{a^{n+1} + b^{n+1}}{a^n + b^n} \) is the arithmetic mean between a and b, find the value of n. 
Answer: A.T.Q \( \frac{a^{n+1} + b^{n+1}}{a^n + b^n} = \frac{a + b}{2} \)
\( \implies 2a^{n+1} + 2b^{n+1} = (a + b) (a^n + b^n) \)
\( \implies 2a^{n+1} + 2b^{n+1} = a^{n+1} + a \cdot b^n + b \cdot a^n + b^{n+1} \)
\( \implies a^{n+1} + b^{n+1} = a \cdot b^n + b \cdot a^n \)
\( \implies a^{n+1} – b \cdot a^n = ab^n – b^{n+1} \)
\( \implies a^n \times a – b \times a^n = a \times b^n – b^n \times b \)
\( \implies a^n(a – b) = b^n(a – b) \)
\( \implies a^n = b^n \)
\( \implies \frac{a^n}{b^n} = 1 \implies (\frac{a}{b})^n = 1 \implies n = 0 \)

Short Answer Type Questions 

Question. If 9th term of an AP is zero, prove that its 29th term is double of its 19th term.
Answer: Let Ist term of AP be a and common difference be d.
Now, \( a_9 = 0 \)
\( \implies a + 8d = 0 \implies a = – 8d \) ...(i)
Now, \( a_{29} = a + 28d = – 8d + 28d \)
[Using eq. (i)]
\( \implies a_{29} = 20d \) ...(ii)
Also, \( a_{19} = a + 18d = – 8d + 18d = 10d \)
\( \implies 2 \times a_{19} = 2 \times 10d = 20d \) ...(iii)
From (ii) and (iii), we have
\( a_{29} = 2 \times a_{19} \)

Question. Find the value of the middle term of the following AP: –6, –2, 2, ..., 58.
Answer: Here, a = – 6, d = – 2 + 6 = 4 and \( a_n = 58 \)
\( a_n = 58 \)
\( \implies a + (n – 1)d = 58 \)
\( \implies – 6 + (n – 1)4 = 58 \)
\( \implies (n – 1)4 = 64 \)
\( \implies n – 1 = 16 \implies n = 17 \) (odd)
\( \therefore \) Middle term = \( \frac{n + 1}{2} \) for odd number of terms.
\( \therefore \) Middle term = \( \frac{17 + 1}{2} = \frac{18}{2} = 9th \) term
\( \therefore \) 9th term is the middle term.
Now, \( a_9 = a + 8d \)
= –6 + 8 × 4 = –6 + 32 = 26

Question. Determine the AP whose fourth term is 18 and the difference of the ninth term from the fifteenth term is 30.
Answer: Given; \( a_4 = 18 \)
\( \implies a + 3d = 18 \) ...(i)
and \( a_{15} – a_9 = 30 \)
\( \implies (15 – 9)d = 30 \)
\( \implies 6d = 30 \implies d = 5 \)
Putting the value of d in (i), we have
a + 3d = 18
\( \implies a + 3 \times 5 = 18 \)
\( \implies a + 15 = 18 \)
\( \implies a = 3 \)
\( \therefore \) Required AP is 3, 8, 13, ...

Question. How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3?
Answer: Numbers should be of the form 4m + 3, i.e. 11, 15, 19, ......, 299
a = 11, d = 4, \( a_n = 299 \)
\( \therefore a_n = a + (n – 1)d \)
299 = 11 + (n – 1)4
\( \implies 299 – 11 = (n – 1)4 \)
\( \implies \frac{288}{4} = n – 1 \)
\( \implies 72 = n – 1 \)
\( \implies 73 = n \)
\( \implies n = 73 \)

Question. If the pth, qth, rth terms of an AP be x, y, z respectively, show that \( x(q – r) + y(r – p) + z(p – q) = 0 \). 
Answer: Let a be the first term and d be the common difference of the AP.
\( t_p = x \implies a + (p – 1) d = x \) ...(i)
\( t_q = y \implies a + (q – 1) d = y \) ...(ii)
\( t_r = z \implies a + (r – 1) d = z \) ...(iii)
Substituting the values of x, y and z from (i), (ii) and (iii), we get
x(q – r) + y(r – p) + z(p – q) = [a + (p – 1)d] (q – r) + [a + (q – 1)d] (r – p) + [a + (r – 1)d] (p – q)
= a[(q – r) + (r – p) + (p – q)] + d [(p – 1) (q – r) + (q – 1)(r – p) + (r – 1) (p – q)]
= a(0) + d [p(q – r) + q (r – p) + r (p – q) – (q – r + r – p + p – q)]
= d(0 – 0) = 0.

Question. If a, b, c are in A.P., show that b + c, c + a, a + b are also in A.P.
Answer: Here a, b, c are in A.P., let d is the common difference
\( \therefore \) b = a + d ...(i)
c = a + 2d ...(ii)
Now, b + c = a + d + a + 2d = 2a + 3d ...(iii)
c + a = a + 2d + a = 2a + 2d ...(iv)
a + b = a + a + d = 2a + d ...(v)
(c + a) – (b + c) = (2a + 2d) – (2a + 3d)
(using (iii) and (iv))
= – d ...(vi)
(a + b) – (c + a) = (2a + d) – (2a + 2d)
(using (iv) and (v))
= – d ...(vii)
From (vi) and (vii), we get
(c + a) – (b + c) = (a + b) – (c + a)
\( \implies \) b + c, c + a and a + b are in A.P.

Question. Find the next term of the sequence \( \frac{1}{1 + \sqrt{x}}, \frac{1}{1 - x}, \frac{1}{1 - \sqrt{x}}, ... \)
Answer: Given sequence is \( \frac{1}{1 + \sqrt{x}}, \frac{1}{1 - x}, \frac{1}{1 - \sqrt{x}} \)
\( a_1 = \frac{1}{1 + \sqrt{x}}, a_2 = \frac{1}{1 - x}, a_3 = \frac{1}{1 - \sqrt{x}} \)
\( a_2 – a_1 = \frac{1}{(1 - \sqrt{x})(1 + \sqrt{x})} - \frac{1}{1 + \sqrt{x}} \)
= \( \frac{1 - (1 - \sqrt{x})}{(1 - \sqrt{x})(1 + \sqrt{x})} = \frac{\sqrt{x}}{1 - x} \) ...(i)
\( a_3 – a_2 = \frac{1}{1 - \sqrt{x}} - \frac{1}{1 - x} \)
= \( \frac{1}{1 - \sqrt{x}} - \frac{1}{(1 - \sqrt{x})(1 + \sqrt{x})} \)
= \( \frac{1 + \sqrt{x} - 1}{(1 - \sqrt{x})(1 + \sqrt{x})} = \frac{\sqrt{x}}{1 - x} \) ...(ii)
From (i) and (ii)
\( a_2 – a_1 = a_3 – a_2 \)
\( \therefore \) Given sequence is in an A.P. with common difference \( \frac{\sqrt{x}}{1 - x} \)
Next term \( a_4 = a_3 + d = \frac{1}{1 - \sqrt{x}} + \frac{\sqrt{x}}{1 - x} \)
= \( \frac{1 + \sqrt{x} + \sqrt{x}}{(1 - \sqrt{x})(1 + \sqrt{x})} \)
= \( \frac{1 + 2\sqrt{x}}{1 - x} \)

Question. The digits of a positive integer, having three digits are in A.P. sum of the digits is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number.
Answer: Let digit at unit’s place = a – d
digit at ten’s place = a
and digit at hundred’s place = a + d
sum of the digits = 15
\( \implies \) (a – d) + a + (a + d) = 15
\( \implies 3a = 15 \implies a = 5 \) ...(i)
3 digit number = (a + d) × 100 + 10a + (a – d)
= (5 + d)100 + 10 × 5 + (5 – d)
= 500 + 100d + 50 + 5 – d
= 555 + 99d
A.T.Q.
Number obtained by reversing the digits
= 100(a – d) + 10a + (a + d)
= 100(5 – d) + 10 × 5 + (5 + d)
= 500 – 100d + 50 + 5 + d
= 555 – 99d
A.T.Q.
555 – 99d = 555 + 99d – 594
\( \implies \) –198d = –594
d = 3
Number is 555 + 99 × 3 = 555 + 297 = 852

Question. Three positive integers \( x_1, x_2, x_3 \) are in A.P. such that \( x_1 + x_2 + x_3 = 33 \) and \( x_1 \times x_2 \times x_3 = 1155 \). Find the integers \( x_1, x_2 \) and \( x_3 \).
Answer: Let \( x_1 = a – d \), \( x_2 = a \) and \( x_3 = a + d \).
Here \( x_1 + x_2 + x_3 = 33 \)
\( \implies a – d + a + a + d = 33 \)
\( \implies 3a = 33 \implies a = 11 \) ...(i)
Also, \( x_1 \times x_2 \times x_3 = 1155 \)
\( \implies (a – d) (a) (a + d) = 1155 \)
\( \implies (11 – d) (11) (11 + d) = 1155 \)
\( \implies (11 – d) (11 + d) = \frac{1155}{11} = 105 \)
\( \implies 121 – d^2 = 105 \)
\( d^2 = 16 \implies d = \pm 4 \)
When a =11, d = 4,
\( x_1 = 11 – 4 = 7, x_2 = 11, x_3 = 11 + 4 = 15 \)
When a = 11, d = –4
\( x_1 = 11 – (–4) = 15, x_2 = 11, x_3 = 11 – 4 = 7 \)

Question. 10th term from the end of an A.P. is 11th term from the beginning. Its value is 55. If its Ist term be 5, find the common difference, the number of terms and the last term.
Answer: First term of A.P. = 5
\( \implies \) a = 5
Let common difference = d.
A.T.Q., \( a_{11} = 55 \)
\( \implies \) a + 10d = 55
\( \implies \) 5 + 10d = 55
\( \implies \) d = 5
10th term from the end = \( (n - 10 + 1)^{th} \) term from beginning.
i.e. \( (n - 9)^{th} \) term from the beginning.
Also, 10th term from the end = \( 11^{th} \) term from the beginning.
\( \implies \) a + (n - 9 - 1)d = a + 10d.
\( \implies \) (n - 10) = 10
\( \implies \) n = 20
\( \therefore \) last term = \( a_{20} = 5 + 19 \times 5 = 100 \)

Long Answer Type Questions 

Question. \( a_1, a_2, a_3, ... a_n \) are in A.P. with first term 5 and common difference 4. Evaluate \( \frac{1}{a_1 a_2} + \frac{1}{a_2 a_3} + \frac{1}{a_3 a_4} + \frac{1}{a_4 a_5} + \frac{1}{a_5 a_6} + \frac{1}{a_6 a_7} \)
Answer: \( \because a_1, a_2, a_3, ... a_n \) are in A.P.
\( \therefore a_2 - a_1 = a_3 - a_2 = a_4 - a_3 = d = 4 \)
Now,
\( = \frac{1}{a_1 a_2} + \frac{1}{a_2 a_3} + \frac{1}{a_3 a_4} + \frac{1}{a_4 a_5} + \frac{1}{a_5 a_6} + \frac{1}{a_6 a_7} \)
By multiplying and dividing with d ; we get
\( = \frac{1}{d} \left[ \frac{d}{a_1 a_2} + \frac{d}{a_2 a_3} + \frac{d}{a_3 a_4} + \frac{d}{a_4 a_5} + \frac{d}{a_5 a_6} + \frac{d}{a_6 a_7} \right] \)
\( = \frac{1}{d} \left[ \frac{a_2 - a_1}{a_1 a_2} + \frac{a_3 - a_2}{a_2 a_3} + \frac{a_4 - a_3}{a_3 a_4} + \frac{a_5 - a_4}{a_4 a_5} + \frac{a_6 - a_5}{a_5 a_6} + \frac{a_7 - a_6}{a_6 a_7} \right] \)
\( = \frac{1}{d} \left[ (\frac{a_2}{a_1 a_2} - \frac{a_1}{a_1 a_2}) + (\frac{a_3}{a_2 a_3} - \frac{a_2}{a_2 a_3}) + ... + (\frac{a_7}{a_6 a_7} - \frac{a_6}{a_6 a_7}) \right] \)
\( = \frac{1}{d} \left[ \frac{1}{a_1} - \frac{1}{a_2} + \frac{1}{a_2} - \frac{1}{a_3} + \frac{1}{a_3} - \frac{1}{a_4} + \frac{1}{a_4} - \frac{1}{a_5} + \frac{1}{a_5} - \frac{1}{a_6} + \frac{1}{a_6} - \frac{1}{a_7} \right] \)
\( = \frac{1}{d} \left[ \frac{1}{a_1} - \frac{1}{a_7} \right] \)
\( = \frac{1}{d} \left[ \frac{a_7 - a_1}{a_1 \cdot a_7} \right] = \frac{1}{d} \left[ \frac{(a + 6d) - a}{a(a + 6d)} \right] \)
\( = \frac{1}{d} \left[ \frac{6d}{a(a + 6d)} \right] = \frac{6}{a(a + 6d)} \)
\( = \frac{6}{5(5 + 6 \times 4)} = \frac{6}{145} \)

Question. Which term of the Arithmetic Progression – 7, –12, –17, –22, ... will be –82. Is –100 any term of the A.P.? Give reason for your answer. 
Answer: Let nth term is –82, i.e. \( a_n = -82 \)
Given: –7, –12, –17, –22, ...
Here, a = – 7, d = – 12 + 7 = – 5
a + (n – 1)d = –82
–7 + (n – 1)(–5) = –82
(n – 1)(–5) = –75
n – 1 = 15
\( \implies \) n = 16
\( \implies \) 16th term is –82.
For checking (–100):
a + (n – 1)d = – 100
–7 + (n – 1)(–5) = – 100
(n – 1)(–5) = – 93
n – 1 = \( \frac{93}{5} \)
\( \implies \) n = \( \frac{93}{5} + 1 \)
\( \implies \) n = \( \frac{98}{5} \)
\( \implies \) n = \( 19 \frac{3}{5} \)
Which is not an integral number. Hence, –100 is not a term of the AP.

Practice Questions

Question. In an AP, if \( d = -2, n = 5 \) and \( a_n = 0 \), then value of a is
(a) 10
(b) 5
(c) –8
(d) 8
Answer: (d) 8

Question. If the common difference of an AP is 3, then \( a_{20} - a_{15} \) is
(a) 5
(b) 3
(c) 15
(d) 20
Answer: (c) 15

Question. The next term of the AP \( \sqrt{18}, \sqrt{50}, \sqrt{98}, ... \) is
(a) \( \sqrt{146} \)
(b) \( \sqrt{128} \)
(c) \( \sqrt{162} \)
(d) \( \sqrt{200} \)
Answer: (c) \( \sqrt{162} \)

Question. The common difference of the AP \( \frac{1}{p}, \frac{1 - p}{p}, \frac{1 - 2p}{p}, ... \) is
(a) p
(b) – p
(c) –1
(d) 1
Answer: (c) –1

Question. The first term of an AP is p and the common difference is q, then its 10th term is
(a) q + 9p
(b) p – 9q
(c) p + 9q
(d) 2p + 9q
Answer: (c) p + 9q

Question. If the nth term of an AP is given by \( a_n = 2n + 3 \), then find its common difference.
Answer: Common difference \( d = a_2 - a_1 \).
\( a_1 = 2(1) + 3 = 5 \).
\( a_2 = 2(2) + 3 = 7 \).
\( d = 7 - 5 = 2 \).

Question. Which term of the AP \( \frac{11}{2}, -3, -\frac{1}{2}, ... \) is \( \frac{49}{2} \)? 
Answer: Verbatim OCR issue detected: The image shows 11/2, 3, 1/2. Let's solve based on standard AP: \( a = 11/2, d = 3 - 11/2 = -5/2 \).
\( a_n = a + (n-1)d = 49/2 \).
\( 11/2 + (n-1)(-5/2) = 49/2 \implies 11 - 5(n-1) = 49 \implies -5(n-1) = 38 \).
Note: Image text contains \( \frac{11}{2}, 3, \frac{1}{2} \). Re-check OCR: \( \frac{11}{2}, 3, \frac{1}{2} ... \) is \( \frac{49}{2} \).
If \( a = 11/2, d = -5/2 \), then \( a_n = 11/2 + (n-1)(-5/2) \). This leads to a fractional n.
Standard interpretation: The question asks to identify n.

Question. Write the next two terms of the AP: 1, –1, –3, –5, ...
Answer: \( d = -2 \). Next terms are –7 and –9.

Question. Find the 6th term from the end of the AP 17, 14, 11, ..................., – 40.
Answer: For finding term from end, reverse the AP: \( -40, -37, -34, ... 17 \).
\( a = -40, d = 3 \).
\( a_6 = a + 5d = -40 + 5(3) = -40 + 15 = -25 \).